On the chromatic number of a random hypergraph

Martin Dyer, Alan Frieze, Catherine Greenhill

Introduction

We study the problem of kk-colouring a random rr-uniform hypergraph with nn vertices and cncn edges, where k, rk,\,r and cc are considered to be constant as n→∞n\to\infty. We generalise a theorem of Achlioptas and Naor for kk-colouring a random graph (22-uniform hypergraph) on nn vertices.

Their theorem specifies the two possible values for the chromatic number of the random graph as n→∞n\to\infty. We give a complete generalisation of the result of . We broadly follow the approach of Achlioptas and Naor , although they rely on simplifications which are available only in the case r=2r=2. We show that these simplifications can be replaced by more general techniques, valid for all k, r≥2k,\,r\geq 2 except k=r=2k=r=2.

There is an extensive literature on this problem in the case r=2r=2, colouring random graphs. In the setting we consider here, this culminates with the results of Achlioptas and Naor , though these do not give a complete answer to the problem. Our results here include those of .

There is also a literature for the case k=2k=2, random hypergraph 22-colouring. Achlioptas, Kim, Krivelevich and Tetali gave a constructive approach, but their results were substantially improved by Achlioptas and Moore , using non-constructive methods. The results of are asymptotic in rr. Our results here include those of , but we also give a non-asymptotic treatment. Recently, Coja-Oghlan and Zdeborová have given a small qualitative improvement of the result of , which goes beyond what can be proved here. See these papers, and their references, for further information.

Finally, we note that Krivelevich and Sudakov studied a wide range of random hypergraph colouring problems, and some of their results were recently improved by Kupavskii and Shabanov . But, in the setting of this paper, these results are much less precise than those we establish here.

After preparing this paper, we learnt of related work by Coja-Oghlan and his coauthors in the case r=2r=2. Coja-Oghlan and Vilenchik improved the upper bound on the kk-colourability threshold, restricting the sharp threshold for kk-colourability to an interval of constant width, compared with logarithmic width in . (See also Remark 3.7 below.) A small improvement in the lower bound was obtained by Coja-Oghlan . Additionally, Coja-Oghlan, Efthymiou and Hetterich adapted the methods from to study kk-colourability of random regular graphs.

Let [n]={1,2,…,n}[n]=\{1,2,\ldots,n\}. Unless otherwise stated, the asymptotic results in this paper are as n→∞n\to\infty. Consider the set Ω(n,r,m)\Omega(n,r,m) of rr-uniform hypergraphs on the vertex set [n][n] with mm edges. Such a hypergraph is defined by its edge set E\mathcal{E}, which consists of mm distinct rr-subsets of nn. Let N=(nr)N=\binom{n}{r} denote the total number of rr-subsets.

Now let G(n,r,m)\mathcal{G}(n,r,m) denote the uniform model of a random rr-regular hypergraph with mm edges. So G(n,r,m)\mathcal{G}(n,r,m) consists of the set Ω(n,r,m)\Omega(n,r,m) equipped with the uniform probability distribution. We write G∈G(n,r,m)G\in\mathcal{G}(n,r,m) for a random hypergraph chosen uniformly from Ω(n,r,m)\Omega(n,r,m). The edge set E\mathcal{E} of this random hypergraph may be viewed as a sample of size mm chosen uniformly, without replacement, from the set of NN possible edges.

Although our main focus is the uniform model G\mathcal{G}, it is simpler for many calculations to work with an alternative model. Let Ω∗(n,r,m)\Omega^{*}(n,r,m) denote the set of all rr-uniform multi-hypergraphs on [n][n], defined as follows: each element of Ω∗(n,r,m)\Omega^{*}(n,r,m) consists of vertex set [n][n] and a multiset of edges, where each edge is now a multiset of rr vertices (not necessarily distinct). We can generate a random element of GG of Ω∗(n,r,m)\Omega^{*}(n,r,m) using the following simple procedure: choose v=(v1,v2,…,vrm)∈[n]rm\boldsymbol{v}=(v_{1},v_{2},\ldots,v_{rm})\in[n]^{rm} uniformly at random and let the edge multiset of GG be {e1,…,em}\{e_{1},\ldots,e_{m}\}, where ei={vr(i−1)+1,…,vri}e_{i}=\{v_{r(i-1)+1},\ldots,v_{ri}\} for i∈[m]i\in[m]. Let G∗(n,r,m)\mathcal{G}^{*}(n,r,m) denote the probability space on Ω∗(n,r,m)\Omega^{*}(n,r,m) which arises from this procedure, and write G∈G∗(n,r,m)G\in\mathcal{G}^{*}(n,r,m) for a hypergraph GG generated in this fashion.

Observe that an element G∈Ω∗(n,r,m)G\in\Omega^{*}(n,r,m) may not satisfy the definition of rr-uniform hypergraph given above, for two reasons. First, an edge of GG may contain repeated vertices, which Ω(n,r,m)\Omega(n,r,m) does not permit. We call such an edge defective. Second, an edge of G∈Ω∗(n,r,m)G\in\Omega^{*}(n,r,m) may be identical to some other edge, which again Ω(n,r,m)\Omega(n,r,m) does not permit. We call such an edge a duplicate.

Say an edge is bad if it is a defective or duplicate edge. Note that G∗(n,r,m)\mathcal{G}^{*}(n,r,m) is not the uniform probability space over Ω∗(n,r,m)\Omega^{*}(n,r,m), but that all GG without bad edges are equiprobable. Thus G∗(n,r,m)\mathcal{G}^{*}(n,r,m), conditional on there being no bad edges, is identical to G(n,r,m)\mathcal{G}(n,r,m).

If EnE_{n} is a sequence of events, we say that EnE_{n} occurs “asymptotically almost surely” (a.a.s.) if Pr⁡(En)→1\Pr(E_{n})\to 1 as n→∞n\to\infty. In this paper, the event EnE_{n} usually concerns G∈G∗(n,r,m)G\in\mathcal{G}^{*}(n,r,m), where m(n)=⌊cn⌋m(n)=\lfloor cn\rfloor, for some constant cc. The difference between cncn and ⌊cn⌋\lfloor cn\rfloor is usually negligible, and we follow in disregarding it unless the distinction is important. Thus we will write the model simply as G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn), and similarly for the other models we consider.

Let cc be a positive constant. For G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn),

Furthermore, for G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn), a.a.s. GG has at most 2ln⁡n2\ln n bad edges.

Throughout this proof, all probabilities are calculated in G∗(n,r,cn)\mathcal{G}^{*}(n,r,cn). For any edge e∈Ee\in\mathcal{E},

Since this is true independently for each e∈Ee\in\mathcal{E}, we have

as m∼cnm\sim cn. Next note that, conditional on there being no defective edges, E\mathcal{E} is a uniform sample of size mm chosen, with replacement, from the NN possible rr-subsets of [n][n]. Thus

Combining (1) and (2) proves the first statement.

Now let mdefm_{\textrm{def}} (mdupm_{\textrm{dup}}, mbadm_{\textrm{bad}}, respectively) denote the number of defective edges (duplicate edges, bad edges, respectively) in G∈G∗(n,r,m)G\in\mathcal{G}^{*}(n,r,m) (counting multiplicities). For the second statement, note that mdefm_{\textrm{def}} has distribution Bin(m,pdefm,p_{\textrm{def}}), and so E[mdef]∼cr(r−1)/2\mathbf{E}[m_{\textrm{def}}]\sim cr(r-1)/2 as n→∞n\to\infty. Hence Chernoff’s bound [13, Corollary 2.4] gives, for large enough nn,

Therefore a.a.s. G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn) has at most ln⁡n\ln n defective edges. Next, note that each edge in E\mathcal{E} has at most (m−1)/nr(m-1)/n^{r} duplicates in expectation, and so

for large nn. (Indeed, if r>2r>2 then E[mdup]=o(1)\mathbf{E}[m_{\textrm{dup}}]=o(1), but we do not exploit this.) Thus, using Markov’s inequality [13, (1.3)],

Combining this with (3) proves the second statement, since mbad≤mdef+mdupm_{\textrm{bad}}\leq m_{\textrm{def}}+m_{\textrm{dup}}. ∎

As already stated, conditional on there being no bad edges, G∗(n,r,cn)\mathcal{G}^{*}(n,r,cn) is identical to G(n,r,cn)\mathcal{G}(n,r,cn). By the first statement of Lemma 1.1, GG has no bad edges with probability Ω(1)\Omega(1) as n→∞n\to\infty. This implies that any event occurring a.a.s. in G∗(n,r,cn)\mathcal{G}^{*}(n,r,cn) occurs a.a.s. in G(n,r,cn)\mathcal{G}(n,r,cn). In Lemma 1.4 we use the second statement of Lemma 1.1 to show that G(n,r,cn)\mathcal{G}(n,r,cn) and G∗(n,r,cn)\mathcal{G}^{*}(n,r,cn) are essentially equivalent, for our purposes.

We also make use of the following simple property of G∗(n,r,m)\mathcal{G}^{*}(n,r,m). A vertex i∈[n]i\in[n] of G∈Ω∗(n,r,m)G\in\Omega^{*}(n,r,m) is isolated if it appears in no edge. Note that a vertex is isolated if and only if it is absent from the vector v∈[n]rm\boldsymbol{v}\in[n]^{rm} defined above. The following simply restates this property.

For S⊆[n]S\subseteq[n], let IS\mathcal{I}_{S} be the event that all vertices in SS are isolated in G∈G∗(n,r,m)G\in\mathcal{G}^{*}(n,r,m). Let G′G^{\prime} be GG conditional on IS\mathcal{I}_{S} and let G′′G^{\prime\prime} be obtained from G′G^{\prime} by deleting all vertices in SS and relabelling the remaining vertices by [n−∣S∣][n-|S|], respecting the original ordering. Then G′′∈G∗(n−∣S∣,r,m)G^{\prime\prime}\in\mathcal{G}^{*}(n-|S|,r,m).

We show, in the proof of Lemma 4.1, that G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn) has Ω(n)\Omega(n) isolated vertices a.a.s. and hence GG has many disconnected components.

A further model of random hypergraphs is often used, which we will denote by G^(n,r,p)\widehat{\mathcal{G}}(n,r,p). In this, the edge set E\mathcal{E} of GG is chosen by Bernoulli sampling. Each of the NN possible rr-subsets of [n][n] is included in E\mathcal{E} independently with probability pp. Essentially, this is G(n,r,m)\mathcal{G}(n,r,m) where mm is a binomial random variable Bin(N,pN,p). We show in Section 1.2 below that G^(n,r,cn/N)\widehat{\mathcal{G}}(n,r,cn/N) and G(n,r,cn)\mathcal{G}(n,r,cn) are equivalent for our problem.

2 Hypergraph colouring

Note that what we study here is sometimes called the weak chromatic number of the hypergraph. The strong chromatic number is defined similarly in terms of strong colourings, which are kk-partitions σ\sigma such that ∣σ(e)∣=∣e∣|\sigma(e)|=|e| for each edge e∈Ee\in\mathcal{E}. Even more general notions of colouring may be defined. See, for example, . We will not consider this further here, though it seems probable that the methods we use would be applicable.

The principal objective of the paper will be to prove the following result.

Define ur,k=kr−1ln⁡ku_{r,k}=k^{r-1}\ln k for integers r≥2r\geq 2 and k≥1k\geq 1. Suppose that r≥2r\geq 2, k≥1k\geq 1, and let cc be a positive constant. Then for G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn),

If c≥ur,kc\geq u_{r,k} then a.a.s. χ(G)>k\chi(G)>k.

If k≥2k\geq 2 and max⁡{r,k}≥3\max\{r,k\}\geq 3 then there exists a constant cr,k∈(ur,k−1, ur,k)c_{r,k}\in(u_{r,k-1},\,u_{r,k}) such that if c<cr,kc<c_{r,k} is a positive constant then a.a.s. χ(G)≤k\chi(G)\leq k.

Now the following theorem, which is a complete generalisation of the result of to uniform hypergraphs, follows easily. Note that the lower bound ur,k−1≤cu_{r,k-1}\leq c from Theorem 1.2 is trivial when k=2k=2, since ur,1=0u_{r,1}=0 for all r≥2r\geq 2.

For all r, k≥2r,\,k\geq 2, if c∈[ur,k−1, ur,k)c\in[u_{r,k-1},\,u_{r,k}) is a positive constant then a.a.s. the chromatic number of G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn) is either kk or k+1k+1. Indeed, if max⁡{r,k}≥3\max\{r,k\}\geq 3 and c∈[ur,k−1, cr,k)c\in[u_{r,k-1},\,c_{r,k}), where cr,kc_{r,k} is a constant satisfying the conditions of Theorem 1.1(b), then a.a.s. χ(G)=k\chi(G)=k for G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn).

Let G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn) and suppose that ur,k−1≤c<ur,ku_{r,k-1}\leq c<u_{r,k}. By Theorem 1.1(a), we know that χ(G)≥k\chi(G)\geq k a.a.s., and by Theorem 1.1(b) we know that χ(G)≤k+1\chi(G)\leq k+1 a.a.s., since c<ur,k<cr,k+1c<u_{r,k}<c_{r,k+1}. This proves the first statement. Furthermore, if max⁡{r,k}≥3\max\{r,k\}\geq 3 and c<cr,kc<c_{r,k} then χ(G)≤k\chi(G)\leq k a.a.s., by Theorem 1.1(b), proving the final statement. ∎

For all but a few small values of (r,k)(r,k) we will see that cr,kc_{r,k} is much closer to ur,ku_{r,k} than to ur,k−1u_{r,k-1}, so that for most values of cc, the chromatic number of G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn) is a.a.s. uniquely determined. For more detail see Remark 3.7.

Although based on a rather simple idea, the second moment method is often very laborious to apply, and our analysis will be no exception.

A kk-partition is called balanced if ⌊n/k⌋≤ni≤⌈n/k⌉\lfloor n/k\rfloor\leq n_{i}\leq\lceil n/k\rceil for i=1,…,ki=1,\ldots,k. A balanced kk-colouring of a HH is a balanced kk-partition which is also a kk-colouring of HH. For convenience, we will assume that kk divides nn, so in a balanced colouring, each colour class has precisely n/kn/k vertices. Since we suppose kk to be constant, the effects of this assumption are asymptotically negligible as n→∞n\to\infty. (This is proved in Lemma 1.4 below.) Following , our analysis will be carried out mainly in terms of balanced colourings. Indeed, we will apply (5) to the random variable ZZ which is the number of balanced kk-colourings (defined formally in Section 2.1).

Clearly, if Z>0Z>0 then a kk-colouring exists. However, the analysis in Section 2 will only allow us to conclude that c<cr,kc<c_{r,k} implies that lim inf⁡n→∞Pr⁡(Z>0)>0\liminf_{n\to\infty}\Pr(Z>0)>0. Thus, we first prove a weaker statement about G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn):

If r, k≥2r,\,k\geq 2 then there exists a constant cr,k∈(ur,k−1,ur,k)c_{r,k}\in(u_{r,k-1},u_{r,k}) such that if c<cr,kc<c_{r,k} is a positive constant then

Then part (b) of Theorem 1.1 will follow from the fact that there is a sharp threshold for kk-colourability of a random hypergraph (see Lemma 1.3, below). Achlioptas and Naor used a result of Achlioptas and Friedgut which established that random graph kk-colourability has a sharp threshold. We will use instead the following, more general, result.

Hatami and Molloy studied the problem of the existence of a homomorphism from a random hypergraph to a fixed hypergraph HH. They used the Bernoulli random hypergraph model G^(n,r,p)\widehat{\mathcal{G}}(n,r,p), defined at the end of Section 1.1.

Given a fixed hypergraph H=([ν],EH)∈Ω∗(ν,r,μ)H=([\nu],\mathcal{E}_{H})\in\Omega^{*}(\nu,r,\mu), Hatami and Molloy considered the threshold pp for the existence of a homomorphism from G=([n],EG)∈G^(n,r,p)G=([n],\mathcal{E}_{G})\in\widehat{\mathcal{G}}(n,r,p) to HH. A homomorphism from GG to HH is a function σ:[n]→[ν]\sigma:[n]\to[\nu] such that σ(e)∈EH\sigma(e)\in\mathcal{E}_{H} for all e∈EGe\in\mathcal{E}_{G}. If H′H^{\prime} is formed from HH by deleting duplicate edges then the homomorphisms from GG to H′H^{\prime} are identical to those from GG to HH, so we may assume that HH has no duplicate edges. A loop in HH is an edge e∈EHe\in\mathcal{E}_{H} for which the underlying set is a singleton. A triangle in HH is a sequence (v1,e1,v2,e2,v3,e3)(v_{1},e_{1},v_{2},e_{2},v_{3},e_{3}) of distinct vertices vi∈[ν]v_{i}\in[\nu] and edges ei∈EHe_{i}\in\mathcal{E}_{H} (i∈i\in), such that v1, v2∈e1v_{1},\,v_{2}\in e_{1}, v2, v3∈e2v_{2},\,v_{3}\in e_{2} and v1, v3∈e3v_{1},\,v_{3}\in e_{3}. The following was proved in (with minor changes of notation):

Let HH be a connected undirected loopless rr-uniform hypergraph with at least one edge. Then the HH-homomorphism problem has a sharp threshold iff (i) r≥3r\geq 3 or (ii) r=2r=2 and HH contains a triangle.

Here a sharp threshold means that there exists a function p(n)p(n) taking values in $forallsufficientlylargefor all sufficiently largensuchthat,forallsuch that, for all0<\varepsilon<1,,G\in\widehat{\mathcal{G}}(n,r,(1-\varepsilon)p)hasahomomorphismtohas a homomorphism toHa.a.s.,anda.a.s., andG\in\widehat{\mathcal{G}}(n,r,(1+\varepsilon)p)hasnohomomorphismtohas no homomorphism toH$ a.a.s.

The property of having an HH-homomorphism is a monotone decreasing property of GG, that is, an HH-homomorphism cannot be destroyed by deleting arbitrary edges of GG. This fact will be used later.

A random hypergraph in G∈G^(n,r,cn/(nr))G\in\widehat{\mathcal{G}}(n,r,cn/\binom{n}{r}) a.a.s. has cn(1+Θ(n−1/4))cn\left(1+\Theta\left(n^{-1/4}\right)\right) edges (see (7), and G∈G^(n,r,cn/(nr))G\in\widehat{\mathcal{G}}(n,r,cn/\binom{n}{r}) is uniformly random conditioned on the number of edges it contains. Hence if an existence problem has a sharp threshold (with respect to pp) for G^(n,r,p)\widehat{\mathcal{G}}(n,r,p) then it has a sharp threshold (with respect to cc) for G(n,r,cn)\mathcal{G}(n,r,cn). In this setting, existence of a sharp threshold means that there exists a function c(n)=Θ(1)c(n)=\Theta(1) such that, for all 0<ε<10<\varepsilon<1, G∈G(n,r,(1−ε)cn)G\in\mathcal{G}(n,r,(1-\varepsilon)cn) has a homomorphism to HH a.a.s., and G∈G(n,r,(1+ε)cn)G\in\mathcal{G}(n,r,(1+\varepsilon)cn) has no homomorphism to HH a.a.s.

Suppose that r,k≥2r,k\geq 2 with max⁡{k,r}≥3\max\{k,r\}\geq 3, and let cc be a positive constant. Then the problem of kk-colouring G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn) has a sharp threshold.

Take K=([k],EK)∈Ω∗(k,r,μ)K=([k],\mathcal{E}_{K})\in\Omega^{*}(k,r,\mu) to be such that EK\mathcal{E}_{K} contains all rr-multisets with elements in [k][k], except for the kk possible loops. Then μ=(k+r−1r)−k\mu=\binom{k+r-1}{r}-k. It is easy to see that the homomorphisms from a graph GG to KK are precisely the kk-colourings of GG. If r=2r=2 and k≥3k\geq 3 then KK contains a triangle. (We may take vi=i(mod3)+1v_{i}=i\pmod{3}+1 and eie_{i} to be an edge with underlying set ∖{i}\setminus\{i\}, for i∈i\in.) Thus it follows from Theorem 1.3 that the problem of kk-colouring G∈G^(n,r,p)G\in\widehat{\mathcal{G}}(n,r,p) has a sharp threshold unless k=r=2k=r=2. Hence, by Observation 1.3, the problem of kk-colouring G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn) has a sharp threshold unless k=r=2k=r=2. ∎

In the excluded case, which is the question of whether a random graph is 22-colourable, it is known that there is no sharp threshold (see [9, Corollary 7]).

We now use Lemma 1.2 to prove the following.

Suppose that k≥2k\geq 2 and max⁡{r,k}≥3\max\{r,k\}\geq 3. Then (b′) implies (b).

From part (b′) of Theorem 1.1, we have a constant cr,k∈(ur,k−1,ur,k)c_{r,k}\in(u_{r,k-1},u_{r,k}) such that for G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn),

whenever c<cr,kc<c_{r,k} is a positive constant. Then Lemma 1.2 implies that the threshold function c(n)c(n) satisfies lim inf⁡n→∞c(n)≥cr,k\liminf_{n\to\infty}c(n)\geq c_{r,k}. Thus for any c<cr,kc<c_{r,k} we have a.a.s. χ(G)≤k\chi(G)\leq k, proving part (b) of Theorem 1.1. ∎

In fact, we will prove an even weaker statement than (b′).

If r, k≥2r,\,k\geq 2 then there exists a constant cr,k∈(ur,k−1,ur,k)c_{r,k}\in(u_{r,k-1},u_{r,k}) such that for any positive constant c<cr,kc<c_{r,k}, the random hypergraph G∈G∗(kt,r,ckt)G\in\mathcal{G}^{*}(kt,r,ckt) satisfies lim inf⁡t→∞Pr⁡(χ(G)≤k)>0\liminf_{t\to\infty}\Pr(\chi(G)\leq k)>0.

Observe that, in addition to restricting nn to multiples of kk, the random hypergraph model for (b′′) is different from that used in (b′). We now show why (b′′) is sufficient.

Let P∗(n,m)=Pr⁡(χ(G)≤k)\textrm{P}^{*}(n,m)=\Pr(\chi(G)\leq k), where G∈G∗(n,r,m)G\in\mathcal{G}^{*}(n,r,m), and let δ(c)=lim inf⁡t→∞P∗(kt,ckt)\delta(c)=\liminf_{t\to\infty}\textrm{P}^{*}(kt,ckt). Then (b′′) is the statement that there exists a constant cr,k∈(ur,k−1,ur,k)c_{r,k}\in(u_{r,k-1},u_{r,k}) such that δ(c)>0\delta(c)>0 for all positive c<cr,kc<c_{r,k}. Assume that (b′′) holds.

Given nn and c<cr,kc<c_{r,k}, let t=⌊n/k⌋t=\lfloor n/k\rfloor and let c′c^{\prime} be such that c<c′<cr,kc<c^{\prime}<c_{r,k}. We show in Lemma 4.1 that G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn) has at least k−1k-1 isolated vertices a.a.s.. Let II be a set of n−kt≤k−1n-kt\leq k-1 isolated vertices in GG, chosen randomly from the set of isolated vertices in GG. Form G′G^{\prime} from GG by deleting the set II of isolated vertices and relabelling the vertices in G′G^{\prime} with [kt][kt], respecting the relative ordering. By symmetry, each set of size n−ktn-kt is equally likely to be the chosen set II. Hence G′∈G∗(kt,r,cn)G^{\prime}\in\mathcal{G}^{*}(kt,r,cn), by Observation 1.1, since GG can be uniquely reconstructed from G′G^{\prime} and II. So P∗(n,cn)=P∗(kt,cn)−o(1)\textrm{P}^{*}(n,cn)=\textrm{P}^{*}(kt,cn)-o(1). Next, if n≥c′k/(c′−c)n\geq c^{\prime}k/(c^{\prime}-c) then c′kt>c′(n−k)≥cnc^{\prime}kt>c^{\prime}(n-k)\geq cn. Therefore, since kk-colourability is a monotone decreasing property (Observation 1.2), it follows that P∗(kt,cn)≥P∗(kt,c′kt)\textrm{P}^{*}(kt,cn)\geq\textrm{P}^{*}(kt,c^{\prime}kt).

Finally, since c′<cr,kc^{\prime}<c_{r,k}, (b′′) implies that P∗(kt,c′kt)>δ(c′)−o(1)\textrm{P}^{*}(kt,c^{\prime}kt)>\delta(c^{\prime})-o(1), with δ(c′)>0\delta(c^{\prime})>0. Hence we have

By Lemma 1.1, a.a.s. G′∈G∗(n,r,c′n)G^{\prime}\in\mathcal{G}^{*}(n,r,c^{\prime}n) has at most 2ln⁡n2\ln n bad edges. Denote the set of bad edges in G′G^{\prime} by B(G′)B(G^{\prime}). Let G′G^{\prime} be a uniformly chosen element of Ω∗(n,r,c′n)\Omega^{*}(n,r,c^{\prime}n) with at most 2ln⁡n2\ln n bad edges, and form the random hypergraph φ(G′)\varphi(G^{\prime}) as follows: delete B(G′)B(G^{\prime}) and a set of (c′−c)n−∣B(G′)∣(c^{\prime}-c)n-|B(G^{\prime})| randomly chosen good edges from G′G^{\prime}. (If nn is sufficiently large then 2ln⁡n≤(c′−c)n2\ln n\leq(c^{\prime}-c)n, making this procedure possible.) The resulting hypergraph φ(G′)\varphi(G^{\prime}) belongs to Ω(n,r,cn)\Omega(n,r,cn), and, by symmetry, it is a uniformly random element of Ω(n,r,cn)\Omega(n,r,cn). That is, that φ(G′)\varphi(G^{\prime}) has the same distribution as G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn) when G′G^{\prime} is chosen uniformly from those elements of Ω∗(n,r,c′n)\Omega^{*}(n,r,c^{\prime}n) with at most 2ln⁡n2\ln n bad edges.

Now choose a constant c′′c^{\prime\prime} with c′<c′′<cr,kc^{\prime}<c^{\prime\prime}<c_{r,k}. Then by (6) applied to c′c^{\prime}, we have P∗(n,c′n)≥δ(c′′)>0\textrm{P}^{*}(n,c^{\prime}n)\geq\delta(c^{\prime\prime})>0. It follows that for G′∈G∗(n,r,c′n)G^{\prime}\in\mathcal{G}^{*}(n,r,c^{\prime}n),

By monotonicity (Observation 1.2), since φ(G′)\varphi(G^{\prime}) has fewer edges than G′G^{\prime}, we conclude that

Hence using the second statement of Lemma 1.1, Pr⁡(χ(G)≤k)≥δ(c′′)−o(1)\Pr(\chi(G)\leq k)\geq\delta(c^{\prime\prime})-o(1) for G∈G(n,r,cn)G\in\mathcal{G}(n,r,cn). This shows that (b′) holds, completing the proof. ∎

The remainder of the paper will be devoted to proving Theorem 1.1, with part (b) weakened to (b′′). First we obtain expressions for E[Z]\mathbf{E}[Z] and E[Z2]\mathbf{E}[Z^{2}] in Sections 2.1 and 2.2, respectively. The expression for E[Z2]\mathbf{E}[Z^{2}] is analysed using Laplace’s method, under the assumption that constants cr,k∈(ur,k−1,ur,k)c_{r,k}\in(u_{r,k-1},u_{r,k}) exist which satisfy some other useful conditions (see Lemma 2.2). This is established in Section 3, completing the proof. Some remarks about asymptotics are made in Section 3.9.

The analysis of Section 3 will require many technical lemmas, some merely verifying inequalities. These inequalities are obvious for large rr and kk but, since rr and kk are constants, we need to establish precise conditions under which they are true. We relegate the proofs of most technical lemmas to the appendix, since they complicate what are fairly natural and straightforward arguments. Therefore, whenever we use a lemma without proof, the proof can be found in the appendix.

To complete this section, we prove the result corresponding to Theorem 1.2 for the Bernoulli random hypergraph model G^(n,r,p)\widehat{\mathcal{G}}(n,r,p). Recall that ur,k=kr−1ln⁡ku_{r,k}=k^{r-1}\ln k and N=(nr)N=\binom{n}{r}.

Let r≥2r\geq 2. Given a positive constant cc, let k(c,r)k(c,r) be the smallest integer kk such that c≤ur,kc\leq u_{r,k}. (Note, ur,k>0u_{r,k}>0 by definition.) If G∈G^(n,r,cn/N)G\in\widehat{\mathcal{G}}(n,r,cn/N) then χ(G)∈{k(c,r), k(c,r)+1}\chi(G)\in\left\{k(c,r),\,k(c,r)+1\right\} a.a.s.

Let G∈G^(n,r,cn/N)G\in\widehat{\mathcal{G}}(n,r,cn/N), and let mm be its (random) number of edges. Then Chernoff’s bound [13, Corollary 2.3] gives

Therefore cn(1−n−\nicefrac14)≤m≤cn(1+n−\nicefrac14)cn(1-n^{-\nicefrac{{1}}{{4}}})\leq m\leq cn(1+n^{-\nicefrac{{1}}{{4}}}) a.a.s., and hence c′n<m<c′′nc^{\prime}n<m<c^{\prime\prime}n a.a.s. for any positive constants c′c^{\prime}, c′′c^{\prime\prime} such that c′<c<c′′c^{\prime}<c<c^{\prime\prime}.

Let k=k(r,c)k=k(r,c), so ur,k−1<c≤ur,ku_{r,k-1}<c\leq u_{r,k}. Choose c′∈(ur,k−1,c)c^{\prime}\in(u_{r,k-1},c), so m>c′nm>c^{\prime}n a.a.s. Now, conditional on m>c′nm>c^{\prime}n, c′>ur,k−1c^{\prime}>u_{r,k-1} implies χ(G)≥k\chi(G)\geq k a.a.s., by Theorem 1.2 and monotonicity (Observation 1.2).

Similarly, choose c′′∈(c,cr,k+1)c^{\prime\prime}\in(c,c_{r,k+1}), so m<c′′nm<c^{\prime\prime}n a.a.s. Then, conditional on m<c′′nm<c^{\prime\prime}n, c′′<cr,k+1c^{\prime\prime}<c_{r,k+1} implies χ(G)≤k+1\chi(G)\leq k+1 a.a.s., by Theorem 1.2 and Observation 1.2. Thus χ(G)∈{k, k+1}\chi(G)\in\left\{k,\,k+1\right\} a.a.s. ∎

We have shown the equivalence of various models for our problem when max⁡{k,r}≥3\max\{k,r\}\geq 3. We note that this equivalence does not hold for the case k=r=2k=r=2, where the non-existence of a 22-colouring is equivalent to the appearance of an odd cycle in a random graph. This is due to the absence of a sharp threshold for this appearance [9, Corollary 7]. Fortunately, this has little impact on our results.

Moment calculations

Let r≥2r\geq 2, k≥1k\geq 1 and recall that ur,k=kr−1ln⁡ku_{r,k}=k^{r-1}\ln k. Suppose that c≥ur,kc\geq u_{r,k} is a positive constant and let G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn). Then a.a.s. χ(G)>k\chi(G)>k.

First suppose that k=1k=1. Since c>0c>0, the hypergraph GG has at least one edge, so χ(G)>1\chi(G)>1 with probability 1.

For the rest of the proof, assume that k≥2k\geq 2. Consider any kk-partition σ∈Πk\sigma\in\Pi_{k} with block sizes nin_{i} (i∈[k])(i\in[k]). Given σ\sigma, a random edge e∈Ee\in\mathcal{E} is monochromatic with probability

using Jensen’s inequality with the convex function xrx^{r}. Since the edges in E\mathcal{E} are chosen independently, the probability that σ\sigma is a kk-colouring of GG is at most (1−1/kr−1)cn(1-1/k^{r-1})^{cn}. Let XX be the number of kk-colourings of GG. Using (5) and the fact that ∣Πk∣=kn|\Pi_{k}|=k^{n}, we conclude that Pr⁡(X>0)≤E[X]≤(k (1−1/kr−1)c)n\Pr(X>0)\leq\mathbf{E}[X]\leq\left(k\,(1-1/k^{r-1})^{c}\right)^{n}. If c≥ur,kc\geq u_{r,k} then c>(kr−1−\nicefrac12)ln⁡kc>(k^{r-1}-\nicefrac{{1}}{{2}})\ln k, and hence

where we have used Lemma 4.7 in the penultimate inequality. It follows that Pr⁡(X>0)→0\Pr(X>0)\to 0 as n→∞n\to\infty when c>ur,kc>u_{r,k}. ∎

We have proved the slightly stronger bound (kr−1−\nicefrac12)ln⁡k(k^{r-1}-\nicefrac{{1}}{{2}})\ln k. This is used in , and noted, but not used, in . Since the difference is small, we mainly use the simpler bound kr−1ln⁡kk^{r-1}\ln k.

In the remainder of the paper, we will assume that kk divides nn, unless stated otherwise. Recall that ZZ is the number of balanced colourings of G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn). Let Ξk\Xi_{k} denote the set of all balanced kk-partitions of [n][n]. For any balanced partition σ∈Ξk\sigma\in\Xi_{k} and any e⊆[n]e\subseteq[n], let Me(σ)M_{e}(\sigma) be the event that ∣σ(e)∣=1|\sigma(e)|=1. If ee is an edge of G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn) then clearly Pr⁡(Me(σ)‾)=1−1/kr−1\Pr(\overline{M_{e}(\sigma)})=1-1/k^{r-1}, and these events are independent for e∈Ee\in\mathcal{E}. Thus, since ∣Ξk∣=n!/((n/k)!)k|\Xi_{k}|=n!/\big((n/k)!\big)^{k},

We have suppressed the discretisation error cn−⌊cn⌋cn-\lfloor cn\rfloor. This would apparently give an additional O(1)O(1) factor in E[Z]\mathbf{E}[Z] here, and in E[Z2]\mathbf{E}[Z^{2}] below. This is of no consequence for two reasons:

We need only prove that lim inf⁡n→∞E[Z2]/E[Z]=Ω(1)\liminf_{n\to\infty}\mathbf{E}[Z^{2}]/\mathbf{E}[Z]=\Omega(1), so the correction is unimportant.

The asymptotic value for E[Z2]/E[Z]\mathbf{E}[Z^{2}]/\mathbf{E}[Z] we obtain is independent of nn, so using the sequence cn=⌊cn⌋/nc_{n}=\lfloor cn\rfloor/n gives the same asymptotic approximation as that given by using cc.

2 Second moment

Using the notation of Section 2.1, let σ,τ∈Ξk\sigma,\tau\in\Xi_{k} be balanced partitions. Then Me(σ)‾∩Me(τ)‾\overline{M_{e}(\sigma)}\cap\overline{M_{e}(\tau)} is the event that the edge ee is not monochromatic in either σ\sigma or τ\tau. For i,j∈[k]i,j\in[k], define

Now Pr⁡(Me(σ))=Pr⁡(Me(τ))=1/kr−1\Pr(M_{e}(\sigma))=\Pr(M_{e}(\tau))=1/k^{r-1}, and

We now apply Stirling’s inequality in the form

Let J0\boldsymbol{J}_{0} be the k×kk\times k matrix with all entries equal to 1/k21/k^{2}. Then

Hence the term of (10) corresponding to L=nJ0\boldsymbol{L}=n\boldsymbol{J}_{0} is asymptotically equal to

Observe from (8) that this term is smaller than E[Z]2\mathbf{E}[Z]^{2} by a factor which is polynomial in nn. We will find a positive constant cr,kc_{r,k} such that when c∈(0,cr,k)c\in(0,c_{r,k}), the function F(X)F(\boldsymbol{X}) has a unique maximum at X=J0\boldsymbol{X}=\boldsymbol{J}_{0}. This will allow us to apply the following theorem of Greenhill, Janson and Ruciński to estimate E[Z2]\mathbf{E}[Z^{2}] in the region where c<cr,kc<c_{r,k}. (See that paper for background and definitions.)

ϕ\phi is twice continuously differentiable in a neighbourhood of x0x_{0} and H:=D2ϕ(x0)H:=D^{2}\phi(x_{0}) is its Hessian at x0x_{0}.

Then provided det⁡(−H∣V)≠0\det(-H|_{V})\neq 0, as n→∞n\to\infty,

and the affine subspace \euW\eu{W} will consist of the matrices X\boldsymbol{X} such that all row and column sums are 1/k1/k, i.e.

The point \euw∈\euW\eu{w}\in\eu{W} will be J0\boldsymbol{J}_{0}.

We wish to calculate E[Z2]\mathbf{E}[Z^{2}], which by (10) equals

In Section 3 we will prove the following result.

Recall that ur,k=kr−1ln⁡ku_{r,k}=k^{r-1}\ln k for r≥2,k≥1r\geq 2,k\geq 1. Now fix r,k≥2r,k\geq 2. There exists a positive constant cr,k∈(ur,k−1, ur,k)c_{r,k}\in(u_{r,k-1},\,u_{r,k}) which satisfies

such that FF has a unique maximum in \euK∩W\eu{K\cap W} at the point J0∈\euK0∩W\boldsymbol{J}_{0}\in\eu{K_{0}\cap W} whenever c∈(0,cr,k)c\in(0,c_{r,k}).

Throughout this section we assume that Lemma 2.2 holds. Then J0\boldsymbol{J}_{0} is the unique maximum of FF within \euK∩W\eu{K\cap W}, so we set \euetahi:=F\euetahi:=F and \eux0:=J0\eu{x_{0}}:=\boldsymbol{J}_{0}. Note that FF is analytic in a neighbourhood of J0\boldsymbol{J}_{0}.

for any k2×(k−1)2k^{2}\times(k-1)^{2} matrix UU whose columns form a basis of M\mathcal{M}.

Suppose that r,k≥2r,k\geq 2 and 0<c<cr,k0<c<c_{r,k}, where cr,kc_{r,k} satisfies Lemma 2.2. Then the determinant of L\mathcal{L} is det⁡L=kk−1\det\mathcal{L}=k^{k-1} and the determinant of −H∣M-H|_{\mathcal{M}} is (k2\eualfa)(k−1)2(k^{2}\eualfa)^{(k-1)^{2}}, where

It follows that MM is a (k−1)×(k−1)(k-1)\times(k-1) block matrix, with blocks of size (k−1)×(k−1)(k-1)\times(k-1), such that

We compute the determinant of matrices of this form in Lemma 4.2. Taking p=q=k−1p=q=k-1 in Lemma 4.2, we have det⁡M=kk−1kk−1=k2(k−1)\det M=k^{k-1}k^{k-1}=k^{2(k-1)}. In particular, since the determinant is nonzero, it follows that the Uij\boldsymbol{U}_{ij} (i,j∈[k−1])(i,j\in[k-1]) give a basis for M\mathcal{M}. Also note that, after permuting its rows,

We also require the determinant of −H∣M-H|_{\mathcal{M}}. For X∈M\boldsymbol{X}\in\mathcal{M}, let

Then H=(hij,i′j′)H=\big(h_{ij,i^{\prime}j^{\prime}}\big) has entries

Here we have used the fact that F2(J0)=(1−1/kr−1)2F_{2}(\boldsymbol{J}_{0})=(1-1/k^{r-1})^{2}.

where IkI_{k} is the k2×k2k^{2}\times k^{2} identity matrix, JJ is the k2×k2k^{2}\times k^{2} matrix with all entries equal to 1,

By (16), the determinant of −H∣M-H|_{\mathcal{M}} equals

Here we have used the fact that JU=0JU=0, which follows since every column of UU is an element of M\mathcal{M} and hence has zero sum. This completes the proof. ∎

Note that \euetasi(J0)=kk2\euetasi(\boldsymbol{J}_{0})=k^{k^{2}}, while (13) gives \euetahi(J0)=F(J0)=2ln⁡(k(1−1/kr−1)c)\euetahi(\boldsymbol{J}_{0})=F(\boldsymbol{J}_{0})=2\ln\big(k(1-1/k^{r-1})^{c}\big). Now \eualfa\eualfa is positive when c∈(0,cr,k)c\in(0,c_{r,k}), using Lemma 2.2. Hence Lemma 2.3 guarantees that det⁡(−H∣M)≠0\det(-H|_{\mathcal{M}})\neq 0. Therefore we can apply Theorem 2.1 to (10), giving

Thus, from (8), for all r, k≥2r,\,k\geq 2 we have

which is a positive constant. So lim inf⁡n→∞Pr⁡(Z>0)>0\liminf_{n\to\infty}\Pr(Z>0)>0 and we have established part (b′′) of Theorem 1.1, under the assumption that Lemma 2.2 holds.

It remains to prove Lemma 2.2, which is the focus of the next section.

Optimisation

We now consider maximising the function FF in (11), and develop conditions under which this function has a unique maximum at J0\boldsymbol{J}_{0}. In doing so, we will determine suitable constants cr,kc_{r,k} and prove that Lemma 2.2 holds. This will complete the proof of Theorem 1.1.

Our initial goal will be to reduce the maximisation of FF to a univariate optimisation problem. This reduction is performed in several stages, presented in Sections 3.2–3.4. We analyse the resulting univariate problem in Sections 3.5–3.8. For more detail on our optimisation strategy, see Section 3.1. Finally, we consider a simplified asymptotic treatment of the univariate optimisation problem in Section 3.9.

As is common when working with convex functions, we define xln⁡x=+∞x\ln x=+\infty for all x<0x<0.

It will be convenient to rescale the variables, letting A=(aij)\boldsymbol{A}=(a_{ij}) be the k×kk\times k matrix defined by A=kX\boldsymbol{A}=k\boldsymbol{X}, so aij=kxija_{ij}=kx_{ij} for all i,j∈[k]i,j\in[k]. Substituting into (11), we can write

Letting z=F(X)−ln⁡kz=F(\boldsymbol{X})-\ln k, we consider the optimisation problem

where we have used Hölder’s inequality in (21). Hence the system (19b)–(19e) is infeasible if ρ∉[1,kr−1]\rho\not\in[1,k^{r-1}], in which case we set max⁡z=−∞\max z=-\infty. Conversely, it is easy to show that the system (19b)–(19e) is feasible for all ρ∈[1,kr−1]\rho\in[1,k^{r-1}]. A formal proof of this is given in Lemma 4.3.

We wish to determine the structure of the maximising solutions in the optimisation problem (19). Following , we will relax the constraints (19d), and write (19b) as

By the same method as for Lemma 4.3, we can show that the system given by (19c), (19e) and (22) is feasible if and only if ∑j=1kϱi=ρ\sum_{j=1}^{k}\varrho_{i}=\rho and ϱi∈[1/k,kr−2]\varrho_{i}\in[1/k,k^{r-2}] for i∈[k]i\in[k]. Note that (20) and (21) assume aij≥0a_{ij}\geq 0, but the relaxation of (19e) will be unimportant. Since z=−∞z=-\infty whenever some aij<0a_{ij}<0, these conditions must be satisfied automatically at any finite optimum.

Having dropped the constraints (19d), we can break (19) up into kk independent simpler one-row subproblems, if we specify the values (ϱ1,ϱ2,…,ϱk)(\varrho_{1},\varrho_{2},\ldots,\varrho_{k}) such that ρ=ϱ1+ϱ2+⋯+ϱk\rho=\varrho_{1}+\varrho_{2}+\cdots+\varrho_{k}. This is done in Section 3.2. Later in Section 3.3 we will determine the optimal values of ϱ1,ϱ2,…,ϱk\varrho_{1},\varrho_{2},\ldots,\varrho_{k} for a fixed value of ρ\rho. Then in Section 3.4 we will allow ρ\rho to vary.

Section 3.2 reduces the one-row problem (23) to an essentially one-variable problem (30). For this problem there is a unique optimum value of β=β(ϱ)\beta=\beta(\varrho) that is given in (29). This value of β\beta maximises the objective function, now expressed as −\euf(β)-\eu{f}(\beta): see (32). The nonlinear constraint (23b) will now have been replaced by an equation \eug(β)=k2−rϱ−k1−r\eu{g}(\beta)=k^{2-r}\varrho-k^{1-r}, see (32).

In Section 3.3 we try to find values ϱ1,ϱ2,…,ϱk\varrho_{1},\varrho_{2},\ldots,\varrho_{k} that sum to a fixed ρ\rho and and associated values β1,β2,…,βk\beta_{1},\beta_{2},\ldots,\beta_{k} that minimise \euf(β1)+\euf(β2)+⋯+\euf(βk)\eu{f}(\beta_{1})+\eu{f}(\beta_{2})+\cdots+\eu{f}(\beta_{k}). The constraints become \eug(β1)+\eug(β2)+⋯+\eug(βk)=k2−r(ρ−1)\eu{g}(\beta_{1})+\eu{g}(\beta_{2})+\cdots+\eu{g}(\beta_{k})=k^{2-r}(\rho-1), see (33). We show that the βi\beta_{i} take one of at most two values, γ1≤γ2\gamma_{1}\leq\gamma_{2}. These values are the solutions to \euf′(β)/\eug′(β)=λ\eu{f}^{\prime}(\beta)/\eu{g}^{\prime}(\beta)=\lambda where λ\lambda is a Lagrange multiplier, to be optimised over. This reduces the optimisation to (42). Here tjt_{j} is the number of βi\beta_{i} taking the value γj\gamma_{j}, and hj=\euf(γj)−λ\eug(γj)h_{j}=\eu{f}(\gamma_{j})-\lambda\eu{g}(\gamma_{j}) for j=1,2j=1,2. We relax the equation in (42) to an inequality and argue that t2=0t_{2}=0 in an optimal solution. At this point we can think of the optimisation as being over λ\lambda or, equivalently, over β∗=γ1\beta_{*}=\gamma_{1}. Choosing the latter we end up with the optimisation problem (44).

We now have to optimise over ρ\rho and find a bound on cc that ensures that aij=1/ka_{ij}=1/k optimises the relaxed problem (19a)–(19c). Using the relaxation (44), we see that it is sufficient to satisfy (46). This leads to an inequality (48) for cc. Making the right hand side of this inequality as small as possible leads to a univariate optimisation problem that is dealt with in Sections 3.5–3.8.

2 The subproblem corresponding to one row

Now, consider any fixed feasible values of the ϱi\varrho_{i} (i∈[k])(i\in[k]) such that ∑j=1kϱi=ρ\sum_{j=1}^{k}\varrho_{i}=\rho. Then the problem decomposes into kk independent maximisation subproblems. In this subsection we use Lagrange multipliers to perform the optimisation on these subproblems.

As already mentioned, when the ϱi\varrho_{i} are fixed (and hence ρ\rho is fixed), the term cln⁡(1−2/kr−1+ρ/k2r−2)c\ln\left(1-2/k^{r-1}+\rho/k^{2r-2}\right) in the objective function zz of (19) is constant. Hence we omit this term from the optimisation problems we consider until we once again allow ρ\rho to vary, in Section 3.4.

We assume that 1/k≤ϱ≤kr−21/k\leq\varrho\leq k^{r-2}, so that the problem is feasible.

When ϱ=1/k\varrho=1/k or ϱ=kr−2\varrho=k^{r-2} the optimization is trivial. If ϱ=1/k\varrho=1/k then there is a unique optimal solution, which satisfies aj=1/ka_{j}=1/k for all j∈[k]j\in[k] and gives z1(ϱ)=ln⁡kz_{1}^{(\varrho)}=\ln k. This is the value of ϱ\varrho that gives the global optimum to our problem. If ϱ=kr−2\varrho=k^{r-2} then there are kk distinct optimal solutions, each with aj=1a_{j}=1 for exactly one value of jj, and aj=0a_{j}=0 otherwise, each giving z1(ϱ)=0z_{1}^{(\varrho)}=0. For ease of exposition, we include these cases in our argument below, though the analysis is unnecessary in these cases.

Introducing the multiplier λ\lambda for (23b) and μ\mu for (23c), the Lagrangian is

The maximisation of LL gives (23b) and (23c), together with the equations

If the equation φ(x)=0\varphi(x)=0 has only one root then all the aja_{j} equal this root and hence, from (23c), aj=1/ka_{j}=1/k for all j∈[k]j\in[k]. In this case ϱ=1/k\varrho=1/k. It follows from Lemma 4.3 that φ\varphi has at least one root.

Now suppose that the equation φ(x)=0\varphi(x)=0 has more than one root (that is, 1/k<ϱ≤kr−21/k<\varrho\leq k^{r-2}), and let α\alpha be the largest. If a\boldsymbol{a} satisfies aj≠αa_{j}\neq\alpha for some j∈[k]j\in[k] then subtracting the corresponding equations in (25) gives

Hence, since −ln⁡x-\ln x and xr−1x^{r-1} are both convex on x>0x>0 and λ\lambda is positive, φ(x)\varphi(x) is a strictly convex function. It follows that the equation φ(x)=0\varphi(x)=0 has at most two roots in (0,∞)(0,\infty). Let the roots of φ(x)=0\varphi(x)=0 be α\alpha and β\beta, where we assume that α>β\alpha>\beta. We have aj∈{α,β}a_{j}\in\left\{\alpha,\beta\right\} for all j∈[k]j\in[k]. But we still need to determine how many of the aja_{j} equal α\alpha and how many equal β\beta.

Consider any stationary point (a∗,λ∗,μ∗)(\boldsymbol{a}_{\ast},\lambda_{\ast},\mu_{\ast}) of LL. Then a∗\boldsymbol{a}_{\ast} and λ∗\lambda_{\ast} satisfy (26). Suppose without loss of generality that for some 1≤t≤k−11\leq t\leq k-1 we have a1,…,at=αa_{1},\ldots,a_{t}=\alpha, at+1,…,ak=βa_{t+1},\ldots,a_{k}=\beta, where a∗=(a1,…,ak)\boldsymbol{a}_{\ast}=(a_{1},\ldots,a_{k}). The Hessian H=Hλ∗,μ∗\boldsymbol{H}=\boldsymbol{H}_{\lambda_{\ast},\mu_{\ast}} of the Lagrangian Lλ∗,μ∗=L( ⋅ ,λ∗,μ∗)L_{\lambda_{\ast},\mu_{\ast}}=L(\,\cdot\,,\lambda_{\ast},\mu_{\ast}), considered as a function of a\boldsymbol{a} only, is a k×kk\times k diagonal matrix with diagonal entries

Since φ\varphi is strictly convex with zeros β<α\beta<\alpha, we know that φ′(β)<0<φ′(α)\varphi^{\prime}(\beta)<0<\varphi^{\prime}(\alpha). The quadratic form determined by the Hessian at a∗\boldsymbol{a}_{\ast} is

To determine the nature of the stationary point a∗\boldsymbol{a}_{\ast}, we restrict the quadratic form to x\boldsymbol{x} lying in the tangent space at a∗\boldsymbol{a}_{\ast}. This means that x\boldsymbol{x} satisfies linear equations determined by the gradient vectors of the constraint functions at a∗\boldsymbol{a}_{\ast}. See, for example, . In our case, these equations are

These equations are linearly independent since α>β\alpha>\beta. Rearranging these equations allows us to express x1x_{1} and xkx_{k} in terms of x2,…,xk−1x_{2},\ldots,x_{k-1} as follows:

For a∗\boldsymbol{a}_{\ast} to be a strict local maximum, the right hand side of (28) must be negative for all x2x_{2}, x3x_{3}, …, xk−1x_{k-1} such that x≠0\boldsymbol{x}\neq\boldsymbol{0}. Since φ′(α)>0\varphi^{\prime}(\alpha)>0, φ′(β)<0\varphi^{\prime}(\beta)<0, this will be true if and only if t=1t=1, when the terms with coefficient φ′(α)\varphi^{\prime}(\alpha) in (28) are absent. This local maximum is clearly unique up to the choice of j∈[k]j\in[k] such that aj=αa_{j}=\alpha. Hence it is global, since z1(ϱ)z_{1}^{(\varrho)} is bounded on the compact region determined by (23b)–(23d), and has no local maxima on the boundary when ϱ<kr−2\varrho<k^{r-2} (see Remark 3.3). Thus there are kk global maxima, given by choosing p∈[k]p\in[k] and setting ap=αa_{p}=\alpha, aj=βa_{j}=\beta (j∈[k],j≠p)(j\in[k],j\neq p), where (α,β)(\alpha,\beta) is the unique solution such that α≥β≥0\alpha\geq\beta\geq 0 to the equations

The fact that there is at least one solution to these equations follows from Lemma 4.3. Next, note that the derivative of the function

is zero at β=1/k\beta=1/k and negative for β∈(0,\nicefrac1k)\beta\in(0,\nicefrac{{1}}{{k}}). Hence there can be at most one solution to these equations which satisfies 0≤β≤\nicefrac1k0\leq\beta\leq\nicefrac{{1}}{{k}}, or equivalently, 0≤β≤α0\leq\beta\leq\alpha.

Note that the relaxation of the constraints (19e) proves to be unimportant, since the optimised values of the aij∈{α,β}a_{ij}\in\{\alpha,\beta\} are positive. Thus the optimisation (23) results in the system

We have omitted the constraint 0≤β0\leq\beta here, but this will be enforced in any optimal solution since z1(ϱ)=−∞z_{1}^{(\varrho)}=-\infty if β<0\beta<0. The maximisation problem is trivial since there is only one feasible solution which satisfies 0≤β≤\nicefrac1k0\leq\beta\leq\nicefrac{{1}}{{k}}, and no other feasible solution can be a maximum.

When ϱ=1/k\varrho=1/k we have α=β=1/k\alpha=\beta=1/k, while if ϱ=kr−2\varrho=k^{r-2} then α=1\alpha=1 and β=0\beta=0. When 1/k<ϱ<kr−21/k<\varrho<k^{r-2} we have 0<β<1/k<α<10<\beta<1/k<\alpha<1.

3 The combined problem, for a fixed value of ρ\rho

We now combine these subproblems (one for each row) to give an optimisation problem corresponding to a fixed value of ρ∈[1,kr−1]\rho\in[1,k^{r-1}], as follows:

As before, the objective function ensures that βi≥0\beta_{i}\geq 0 for i∈[k]i\in[k] at any finite optimum. Recall from Remark 3.1 that z2(ρ)z_{2}^{(\rho)} has no maximum on the boundary when ρ<kr−1\rho<k^{r-1}.

where α\alpha is defined as 1−(k−1)β1-(k-1)\beta and hence dα/dβ=−(k−1)\textrm{d}\alpha/\textrm{d}\beta=-(k-1). We use the notation \euf\eu{f} and \eug\eu{g} here, and reserve the symbols ff and gg for transformed versions of these functions which will be introduced in Section 3.5.

Now \euf(β)=+∞\eu{f}(\beta)=+\infty if β<0\beta<0 or β>1/(k−1)\beta>1/(k-1). Also

so both \euf(β)\eu{f}(\beta) and \eug(β)\eu{g}(\beta) are positive and decreasing for β∈[0,\nicefrac1k)\beta\in[0,\nicefrac{{1}}{{k}}).

Letting z^2(ρ)=kln⁡k−z2(ρ)\widehat{z}_{2}^{(\rho)}=k\ln k-z_{2}^{(\rho)}, (31) can now be rewritten as

We therefore ignore β=0\boldsymbol{\beta}=\boldsymbol{0} in our search for the minimum in (33): see Remark 3.2.

We also ignore (33c) and apply the Lagrangian method to (33a) and (33b), using the multiplier −λ-\lambda for (33b). The Lagrangian optimisation will be to minimise the function

The stationary points of the Lagrangian ψ(ρ)\psi^{(\rho)} are given by (31b), (31c) and the equations

We will concentrate on those stationary points of the Lagrangian that can give us the optimum for (33).

Let B=[0,\nicefrac1k]B=[0,\nicefrac{{1}}{{k}}]. We define

for β∈[0,\nicefrac1k)\beta\in[0,\nicefrac{{1}}{{k}}), and extend by continuity to give

Again, we reserve the notation η\eta and ω\omega for transformed versions of these functions, introduced in Section 3.5. (The values of \eueta(\nicefrac1k)\eueta(\nicefrac{{1}}{{k}}) and \euomega(\nicefrac1k)\euomega(\nicefrac{{1}}{{k}}) are established in Lemma 4.10, in terms of the transformed functions.)

Now suppose that (β∗,λ∗)(\boldsymbol{\beta}_{*},\lambda_{*}) is a stationary point of the Lagrangian ψ(ρ)\psi^{(\rho)} which satisfies β∗=(β1∗,…,βk∗)∈Bk\boldsymbol{\beta}_{*}=(\beta_{1*},\ldots,\beta_{k*})\in B^{k}. Then z^2(ρ)(β∗)=ψ(ρ)(β∗,λ∗)\widehat{z}_{2}^{(\rho)}(\boldsymbol{\beta}_{*})=\psi^{(\rho)}(\boldsymbol{\beta}_{*},\lambda_{*}).

Suppose that there exists ii such that 0<βi∗<1/k\boldsymbol{0}<\beta_{i*}<1/k. Then βi∗<1/k<αi∗=1−(k−1)βi∗\beta_{i*}<1/k<\alpha_{i*}=1-(k-1)\beta_{i*}, and (β,λ)=(βi∗,λ∗)(\beta,\lambda)=(\beta_{i*},\lambda_{*}) must satisfy the equation

where α=1−(k−1)β\alpha=1-(k-1)\beta. This shows that 0<λ∗=\euomega(β∗)<∞0<\lambda_{*}=\euomega(\boldsymbol{\beta}_{*})<\infty. Furthermore, (37) implies that βj∗>0\beta_{j*}>0 for all jj, since \euomega(0)=∞\euomega(0)=\infty. Thus in any stationary point (β∗,λ∗)(\boldsymbol{\beta}_{*},\lambda_{*}) of ψ(ρ)\psi^{(\rho)} with 0≠β∗∈Bk\boldsymbol{0}\neq\boldsymbol{\beta}_{*}\in B^{k}, for each i∈[k]i\in[k], either βi∗=1/k\beta_{i*}=1/k (in which case αi∗=1/k\alpha_{i*}=1/k and (36) holds), or βi∗∈(0,\nicefrac1k)\beta_{i*}\in(0,\nicefrac{{1}}{{k}}) and (βi∗,λ∗)(\beta_{i*},\lambda_{*}) is a solution to (37).

We now assume that β∗≠(\nicefrac1k,…,\nicefrac1k)\boldsymbol{\beta}_{*}\neq(\nicefrac{{1}}{{k}},\ldots,\nicefrac{{1}}{{k}}) (see Remark 3.2) and rewrite (35) as

First suppose that λ∗>max⁡B \eueta(β)\lambda_{*}>\max_{B}\,\eueta(\beta). Since

for all β∈(0,\nicefrac1k)\beta\in(0,\nicefrac{{1}}{{k}}), it follows from (37) that \eueta′(βi∗)<0\eueta^{\prime}(\beta_{i*})<0 for any i∈[k]i\in[k] with βi∗≠\nicefrac1k\beta_{i*}\neq\nicefrac{{1}}{{k}}. But this shows that (β∗,λ∗)(\boldsymbol{\beta}_{*},\lambda_{*}) is not a local minimum of ψ(ρ)\psi^{(\rho)}, as we can decrease the value of ψ(ρ)\psi^{(\rho)} by increasing βi∗\beta_{i*} infinitesimally, while holding all other values of βj∗\beta_{j*} and λ∗\lambda_{*} steady. Since we wish to minimise ψ(ρ)\psi^{(\rho)} (and hence z^2(ρ)\widehat{z}_{2}^{(\rho)}), we now assume that λ∗≤max⁡B \eueta(β)\lambda_{*}\leq\max_{B}\,\eueta(\beta).

Next, suppose that λ∗<min⁡B \eueta(β)\lambda_{*}<\min_{B}\,\eueta(\beta). As β∗≠(\nicefrac1k,…,\nicefrac1k)\boldsymbol{\beta}_{*}\neq(\nicefrac{{1}}{{k}},\ldots,\nicefrac{{1}}{{k}}), by (38) we conclude that

Hence (β∗,λ∗)(\boldsymbol{\beta}_{*},\lambda_{*}) cannot minimise z^2(ρ)\widehat{z}_{2}^{(\rho)} if λ∗<min⁡β∈B\eueta(β)\lambda_{*}<\min_{\beta\in B}\eueta(\beta).

Therefore (see Remark 3.2) we may now assume that (β∗,λ∗)(\boldsymbol{\beta}_{*},\lambda_{*}) is a stationary point of (33) with

where λ∗=\euomega(βi∗)\lambda_{*}=\euomega(\beta_{i*}) for any ii such that βi∗≠\nicefrac1k\beta_{i*}\neq\nicefrac{{1}}{{k}}, and such that λ=λ∗\lambda=\lambda_{*} satisfies

We will prove the following in Section 3.5 below.

The function \euomega(β)\euomega(\beta) has a unique minimum in (0,\nicefrac1k)(0,\nicefrac{{1}}{{k}}). Furthermore, if (39) holds for some λ>0\lambda>0 then the equation \euomega(β)=λ\euomega(\beta)=\lambda has at most two distinct solutions β∈B\beta\in B.

Now consider the case that the equation \euomega(β)=λ\euomega(\beta)=\lambda has exactly two distinct roots γ1>γ2\gamma_{1}>\gamma_{2}. Define γ0=\nicefrac1k\gamma_{0}=\nicefrac{{1}}{{k}}. Let tit_{i} (i=1,2)(i=1,2) be the multiplicity of γi\gamma_{i} amongst the βj\beta_{j} (j∈[k]j\in[k]). We write \eufi\eu{f}_{i} for \euf(γi)\eu{f}(\gamma_{i}) (i=0,1,2)(i=0,1,2), and similarly for \eug\eu{g}, \eueta\eueta and \euomega\euomega. For i=0,1,2i=0,1,2 we define hi=\eufi−λ\eugih_{i}=\eu{f}_{i}-\lambda\eu{g}_{i}. Since γ1>γ2\gamma_{1}>\gamma_{2} and \eug′(β)<0\eu{g}^{\prime}(\beta)<0 for β∈[0,\nicefrac1k)\beta\in[0,\nicefrac{{1}}{{k}}), we have \eug1<\eug2\eu{g}_{1}<\eu{g}_{2}. Now \euomega\euomega is continuous on BB and has a unique minimum in (0,\nicefrac1k)(0,\nicefrac{{1}}{{k}}), by Lemma 3.1. Hence, for any β\beta strictly between the two solutions γ1,γ2\gamma_{1},\gamma_{2} of \euomega(β)=λ\euomega(\beta)=\lambda, it follows that \euomega(β)<λ\euomega(\beta)<\lambda. Therefore

Hence h1<h2h_{1}<h_{2}. Also, as \euf0=\eug0=0\eu{f}_{0}=\eu{g}_{0}=0 we have

where the final inequality holds since \euomega(β)>λ\euomega(\beta)>\lambda for γ1<β<\nicefrac1k\gamma_{1}<\beta<\nicefrac{{1}}{{k}}, by Lemma 3.1 (noting that γ1\gamma_{1} is the larger of the two solutions of \euomega(β)=λ\euomega(\beta)=\lambda). Hence h1<0h_{1}<0. Now the minimum of (33) is bounded below by the solution of the following problem:

We relax the equality constraint in (42) to give

It follows that we must have t2=0t_{2}=0 in the optimal solution to (43). To see this, suppose the optimal solution is t1=τ1t_{1}=\tau_{1}, t2=τ2>0t_{2}=\tau_{2}>0. Consider the solution t1=τ1+τ2t_{1}=\tau_{1}+\tau_{2}, t2=0t_{2}=0. This clearly satisfies the second and third constraint of (43). Since \eug\eu{g} is decreasing on [0,\nicefrac1k][0,\nicefrac{{1}}{{k}}] we have \eug1<\eug2\eu{g}_{1}<\eu{g}_{2}, so

Hence the solution t1=τ1+τ2t_{1}=\tau_{1}+\tau_{2}, t2=0t_{2}=0 also satisfies the first constraint. Now (τ1+τ2)h1<τ1h1+τ2h2(\tau_{1}+\tau_{2})h_{1}<\tau_{1}h_{1}+\tau_{2}h_{2} by (40), contradicting the optimality of t1=τ1,t2=τ2t_{1}=\tau_{1},t_{2}=\tau_{2}. Therefore, we will simply write β∗\beta_{\ast} for γ1\gamma_{1} and tt for t1t_{1} from this point.

The rest of the argument also holds when \euomega(β)=λ\euomega(\beta)=\lambda has only one solution β∗\beta_{\ast}, so this case re-enters the argument now. By (41), we must choose tt to be as large as possible subject to the constraints t≤kt\leq k and t \eug1≤k2−r(ρ−1)t\,\eu{g}_{1}\leq k^{2-r}(\rho-1). Therefore tt must be the smaller of ⌊k2−r(ρ−1)/\eug(β∗)⌋\lfloor k^{2-r}(\rho-1)/\eu{g}(\beta_{\ast})\rfloor and kk. We will usually relax the constraint t≤kt\leq k below, since we are mainly interested in small values of tt. In any case, this relaxation can only worsen the objective function. Recalling that z2(ρ)=kln⁡k−z^2(ρ)z_{2}^{(\rho)}=k\ln k-\widehat{z}_{2}^{(\rho)}, the objective function of the system (31) can be bounded above by

Note that λ\lambda has now been removed from consideration, as this upper bound on z2(ρ)z_{2}^{(\rho)} depends on β∗=β∗(ρ)\beta_{\ast}=\beta_{\ast}(\rho) only.

4 Allowing ρ\rho to vary

Here we return to the relaxed problem given by (19–19b), now allowing the value of ρ\rho to vary.

As ρ\rho increases from 1 to kr−1k^{r-1}, the bound in (44) changes only at integral values of k2−r(ρ−1)/\eug(β∗)k^{2-r}(\rho-1)/\eu{g}(\beta_{\ast}). Thus the only relevant values of ρ\rho of are those for which k2−r(ρ−1)/\eug(β∗)k^{2-r}(\rho-1)/\eu{g}(\beta_{\ast}) is an integer. Then we may write (44) simply as

Let J\boldsymbol{J} be the k×kk\times k matrix with all entries \nicefrac1k\nicefrac{{1}}{{k}}, and note that J/k=J0\boldsymbol{J}/k=\boldsymbol{J}_{0}. We wish to find conditions on cc which guarantee that F(A/k)<F(J/k)F(\boldsymbol{A}/k)<F(\boldsymbol{J}/k) for all A≠J\boldsymbol{A}\neq\boldsymbol{J} which satisfy (19b), (19c). From the above, and (19), this will be true when

Next, from (45) we have (ρ−1)=t \eug(β∗)kr−2(\rho-1)=t\,\eu{g}(\beta_{\ast})k^{r-2}. Substituting this into (47) gives

Now the right side of (50) is clearly minimised when ϑ\vartheta is as small as possible. From (49), this is when tt is as small as possible. If we relax the integrality constraint on tt and allow t→0t\to 0, then ϑ→0\vartheta\to 0 and (50) becomes c<C(β∗)c<C(\beta_{\ast}). Thus we can estimate cr,kc_{r,k} by minimising C(β)C(\beta) over β∈[0,\nicefrac1k]\beta\in[0,\nicefrac{{1}}{{k}}]. Then the computation of cr,kc_{r,k} reduces to minimising the function

Then, whenever c∈(0,cr,k)c\in(0,c_{r,k}), we know that (50) holds, and hence that J\boldsymbol{J} is the unique maximum of FF over all doubly stochastic matrices.

We could use (53) directly to improve the estimate of cr,kc_{r,k}. This is done in for k=2k=2, giving a small improvement in cr,2c_{r,2}, though uses only (52) for r=2r=2. In the main, we will also use (52), which corresponds to allowing t→0t\to 0. However, we show in Section 3.9 that the increment in cr,kc_{r,k} which results from using (53) is small, and can be obtained indirectly from (50).

We might improve the estimate of cr,kc_{r,k} further by avoiding the relaxation of (19d) in the optimisation. We note that taking t=kt=k in (48) results in a local maximum of (19), as follows. Let pp be any permutation of [k][k], and set aip(i)=αa_{ip(i)}=\alpha, aij=βa_{ij}=\beta (j≠p(i), i∈[k]j\neq p(i),\,i\in[k]). This gives k!k! local maxima of (19). We conjecture that these solutions are the global maxima, but we are unable to prove this. The inclusion of (19d) gives conditions for the local maxima which may have solutions yielding larger values of zz in (19). The local maxima seem rather difficult to describe explicitly, so we leave this as an open question. However, we show in Section 3.9 that including (19d) cannot result in a large improvement in cr,kc_{r,k}.

5 The univariate optimisation

We have now achieved the objective of reducing the problem to a univariate optimisation, namely, minimising the function \eueta\eueta. To carry out this minimisation, we will first make a substitution x=(k−1)βx=(k-1)\beta in (51), so that

Figure 1 gives a plot of the function η\eta when k=4k=4 and r=3r=3.

for x∈(0,1−\nicefrac1k)x\in(0,1-\nicefrac{{1}}{{k}}), and at the boundaries we have f(0)=ln⁡kf(0)=\ln k and f(1−\nicefrac1k)=0f(1-\nicefrac{{1}}{{k}})=0. Differentiating gives

Therefore f(x)>f(1−\nicefrac1k)=0f(x)>f(1-\nicefrac{{1}}{{k}})=0 for all x∈(0,1−\nicefrac1k)x\in(0,1-\nicefrac{{1}}{{k}}). Also lim⁡x→0f′(x)=−∞\lim_{x\to 0}f^{\prime}(x)=-\infty while f′(1−\nicefrac1k)=0f^{\prime}(1-\nicefrac{{1}}{{k}})=0. Note, using (55), that

We note that f′′(1−\nicefrac1k)=k2/(k−1)f^{\prime\prime}(1-\nicefrac{{1}}{{k}})=k^{2}/(k-1) and f′′′(1−\nicefrac1k)=k3(k−2)/(k−1)2f^{\prime\prime\prime}(1-\nicefrac{{1}}{{k}})=k^{3}(k-2)/(k-1)^{2}. Now we turn our attention to the function gg, which satisfies g(0)=1−1/kr−1g(0)=1-1/k^{r-1} and g(1−\nicefrac1k)=0g(1-\nicefrac{{1}}{{k}})=0. Differentiating gives

Hence f(x)f(x) and g(x)g(x) are positive, strictly decreasing and strictly convex functions on (0,1−\nicefrac1k)(0,1-\nicefrac{{1}}{{k}}).

Returning to the function η\eta defined in (54), in Lemma 4.10 we show that

and we will take these limits as defining η(1−\nicefrac1k)\eta(1-\nicefrac{{1}}{{k}}), η′(0)\eta^{\prime}(0) and η′(1−\nicefrac1k)\eta^{\prime}(1-\nicefrac{{1}}{{k}}), respectively. Note also that η(0)=kr−1ln⁡k/(kr−1−1)\eta(0)=k^{r-1}\ln k/(k^{r-1}-1).

If k=2k=2 then η\eta has a stationary point at x=1−\nicefrac1k=\nicefrac12x=1-\nicefrac{{1}}{{k}}=\nicefrac{{1}}{{2}}. Otherwise, η\eta has an interior minimum in (0,1−\nicefrac1k)(0,1-\nicefrac{{1}}{{k}}), since η′(0)<0\eta^{\prime}(0)<0 and η′(1−\nicefrac1k)>0\eta^{\prime}(1-\nicefrac{{1}}{{k}})>0. We first show that this is the unique stationary point of η\eta in (0,1−\nicefrac1k)(0,1-\nicefrac{{1}}{{k}}). This is not straightforward, since η\eta is not convex, as observed in for the case r=2r=2. Furthermore, the approach of , making a nonlinear substitution in η\eta, does not generalise beyond r=2r=2. Hence our arguments here are very different from those in .

To determine the nature of the stationary points of η\eta, we consider the function h(x)=f(x)−λg(x)h(x)=f(x)-\lambda g(x) on (0,1−\nicefrac1k](0,1-\nicefrac{{1}}{{k}}], for fixed λ>0\lambda>0. Then hh is analytic, and its zeros contain the points at which η(x)=λ\eta(x)=\lambda in (0,1−\nicefrac1k](0,1-\nicefrac{{1}}{{k}}]. We will apply Rolle’s Theorem to hh. The zeros of hh are separated by zeros of h′h^{\prime}, and these are separated by zeros of h′′h^{\prime\prime}. Since f(1−\nicefrac1k)=g(1−\nicefrac1k)=0f(1-\nicefrac{{1}}{{k}})=g(1-\nicefrac{{1}}{{k}})=0 and f′(1−\nicefrac1k)=g′(1−\nicefrac1k)=0f^{\prime}(1-\nicefrac{{1}}{{k}})=g^{\prime}(1-\nicefrac{{1}}{{k}})=0, we conclude that h′h^{\prime} has a zero at x=1−\nicefrac1kx=1-\nicefrac{{1}}{{k}} for all λ\lambda, and hh has a double zero at x=1−\nicefrac1kx=1-\nicefrac{{1}}{{k}}. Now, from (57) and (59), the zeros of h′′(x)=f′′(x)−λg′′(x)h^{\prime\prime}(x)=f^{\prime\prime}(x)-\lambda g^{\prime\prime}(x) in (0,1−\nicefrac1k](0,1-\nicefrac{{1}}{{k}}] are the solutions of

In Lemma 4.11 we show that if r≤2kr\leq 2k then (61) has at most two solutions in $,whileif, while ifr\geq 2k+1then(61)hasatmosttwosolutionsinthen (61) has at most two solutions in[0,1-\nicefrac{{1}}{{k}}]wheneverwhenever\lambda<\lambda_{0}$, where

(Here, as elsewhere in the paper, we have r,k≥2r,k\geq 2.) For uniformity, we set λ0=∞\lambda_{0}=\infty if r≤2kr\leq 2k and define

Then Λ′\Lambda^{\prime} is a union of open intervals. We show in Lemma 4.13 that η(0)<η(1−\nicefrac1k)<λ0\eta(0)<\eta(1-\nicefrac{{1}}{{k}})<\lambda_{0}, which implies that 0, 1−\nicefrac1k∈Λ0,\,1-\nicefrac{{1}}{{k}}\in\Lambda. Hence Λ=Λ′∪{0,1−\nicefrac1k}\Lambda=\Lambda^{\prime}\cup\left\{0,1-\nicefrac{{1}}{{k}}\right\}, which shows that Λ\Lambda is nonempty. Now η′(0)<0, η′(1−\nicefrac1k)≥0\eta^{\prime}(0)<0,\,\eta^{\prime}(1-\nicefrac{{1}}{{k}})\geq 0 imply that Λ′\Lambda^{\prime} is nonempty. Our search for a value of xx making η\eta small will be restricted to Λ′\Lambda^{\prime}. We have shown that h′′h^{\prime\prime} has at most two zeros in Λ\Lambda, and hence hh has at most four zeros in Λ\Lambda. Since there is a double zero of hh at x=1−\nicefrac1k∈Λ∖Λ′x=1-\nicefrac{{1}}{{k}}\in\Lambda\setminus\Lambda^{\prime}, it follows that there are at most two zeros of hh in Λ′\Lambda^{\prime}. Thus η(x)=λ\eta(x)=\lambda at most twice in Λ′\Lambda^{\prime}. Since η′(0)<0, η′(1−\nicefrac1k)≥0\eta^{\prime}(0)<0,\,\eta^{\prime}(1-\nicefrac{{1}}{{k}})\geq 0, we know that η\eta has a local minimum in Λ′\Lambda^{\prime}. Then η\eta has at most one local minimum ξ∈Λ′\xi\in\Lambda^{\prime}. To see this, suppose there are two local minima ξ1, ξ2∈Λ′\xi_{1},\,\xi_{2}\in\Lambda^{\prime} with η(ξ1)≤η(ξ2)=λ<λ0\eta(\xi_{1})\leq\eta(\xi_{2})=\lambda<\lambda_{0}. If η(ξ1)=λ\eta(\xi_{1})=\lambda then η(x)=λ\eta(x)=\lambda has at least four roots in Λ′\Lambda^{\prime}, with double roots at both ξ1\xi_{1} and ξ2\xi_{2}. If η(ξ1)<λ\eta(\xi_{1})<\lambda then η(x)=λ\eta(x)=\lambda has at least three roots in Λ′\Lambda^{\prime}, with a double root at ξ2\xi_{2} and, by continuity, a root strictly between ξ1\xi_{1} and ξ2\xi_{2}. In either case, we have a contradiction. It also follows that Λ\Lambda is connected. Otherwise, since η′(0)<0, η′(1−\nicefrac1k)≥0\eta^{\prime}(0)<0,\,\eta^{\prime}(1-\nicefrac{{1}}{{k}})\geq 0, each maximal interval of Λ′\Lambda^{\prime} must contain a local minimum, a contradiction. Thus Λ=[0,1−\nicefrac1k]\Lambda=[0,1-\nicefrac{{1}}{{k}}]. In other words,

We have proved that η\eta has exactly one local minimum in (0,1−\nicefrac1k)(0,1-\nicefrac{{1}}{{k}}), and we will denote this minimum point by ξ∈(0,1−\nicefrac1k)\xi\in(0,1-\nicefrac{{1}}{{k}}). It also follows that there are no local maxima of η\eta in [0,1−\nicefrac1k][0,1-\nicefrac{{1}}{{k}}], as we now prove. If there were a local maximum ξ′∈[0,ξ)\xi^{\prime}\in[0,\xi) then η′(0)<0\eta^{\prime}(0)<0 would imply that there is a local minimum in (0,ξ′)(0,\xi^{\prime}), a contradiction. The same argument applies to the interval (ξ,1−\nicefrac1k](\xi,1-\nicefrac{{1}}{{k}}], for k>2k>2. If k=2k=2, it is possible that x=\nicefrac12x=\nicefrac{{1}}{{2}} is a local maximum, but it still follows that there can be no local maximum in (ξ,\nicefrac12)(\xi,\nicefrac{{1}}{{2}}).

To summarise: if k>2k>2 then η\eta has exactly one stationary point ξ∈(0,1−\nicefrac1k)\xi\in(0,1-\nicefrac{{1}}{{k}}), a local minimum. If k=2k=2 then there is a unique local minimum ξ∈(0,\nicefrac12]\xi\in(0,\nicefrac{{1}}{{2}}] but, if ξ≠\nicefrac12\xi\neq\nicefrac{{1}}{{2}}, then \nicefrac12\nicefrac{{1}}{{2}} may be a local maximum. In either case, ξ\xi is the global minimum.

We now prove Lemma 3.1, using the same method but working with the transformed function ω\omega defined by

Figure 2 gives a plot of the function ω\omega when k=4k=4 and r=3r=3.

Let λ\lambda be a real number which satisfies (39). Note that the solutions to \euomega(β)=λ\euomega(\beta)=\lambda in (37) correspond to the zeros of h′(x)h^{\prime}(x), where h(x)h(x) is the function defined above. Combining (39) and (62), we see that λ<λ0\lambda<\lambda_{0}. Therefore by Lemma 4.11 and Lemma 4.13, we may conclude that h′′(x)h^{\prime\prime}(x) has at most two zeros in [0,1−\nicefrac1k][0,1-\nicefrac{{1}}{{k}}]. Hence h′(x)h^{\prime}(x) has at most three zeros, and we know that h′(1−\nicefrac1k)=0h^{\prime}(1-\nicefrac{{1}}{{k}})=0. Thus there can be at most two zeros of h′(x)h^{\prime}(x) in [0,1−\nicefrac1k)[0,1-\nicefrac{{1}}{{k}}). Therefore ω(x)=f′(x)/g′(x)\omega(x)=f^{\prime}(x)/g^{\prime}(x) can take the value λ\lambda at most twice in [0,1−\nicefrac1k)[0,1-\nicefrac{{1}}{{k}}). Since ω\omega is analytic on (0,1−\nicefrac1k)(0,1-\nicefrac{{1}}{{k}}), by the arguments above, ω\omega can have at most one stationary point in (0,1−\nicefrac1k)(0,1-\nicefrac{{1}}{{k}}).

Now ω(0)=+∞\omega(0)=+\infty since f′(0)=−∞f^{\prime}(0)=-\infty and g′(0)=−rg^{\prime}(0)=-r. By Lemma 4.10 ω(1−\nicefrac1k)=η(1−\nicefrac1k)=kr−1/r(r−1)\omega(1-\nicefrac{{1}}{{k}})=\eta(1-\nicefrac{{1}}{{k}})=k^{r-1}/r(r-1) and that ω(ξ)=η(ξ)\omega(\xi)=\eta(\xi), where ξ\xi denotes the point which minimises η\eta. Since ω(ξ)=η(ξ)<+∞=ω(0)\omega(\xi)=\eta(\xi)<+\infty=\omega(0) and ω(ξ)=η(ξ)<η(1−\nicefrac1k)=ω(1−\nicefrac1k)\omega(\xi)=\eta(\xi)<\eta(1-\nicefrac{{1}}{{k}})=\omega(1-\nicefrac{{1}}{{k}}), ω\omega must have a unique minimum in (0,1−\nicefrac1k)(0,1-\nicefrac{{1}}{{k}}), completing the proof. ∎

It remains to identify the local minimum ξ\xi of η\eta to a close enough approximation. Using (56), the condition that η′(x)≤0\eta^{\prime}(x)\leq 0 is

We have shown that f′(x)<0f^{\prime}(x)<0 and g′(x)<0g^{\prime}(x)<0 for x∈(0,1−\nicefrac1k)x\in(0,1-\nicefrac{{1}}{{k}}), so the condition η′(x)≤0\eta^{\prime}(x)\leq 0 is equivalent to

We will now use (63) to show that ξ\xi is approximately 1/kr−11/k^{r-1}, except for the cases k=2, r=3, 4k=2,\,r=3,\,4. (If r=2r=2 then ξ=1/kr−1\xi=1/k^{r-1} exactly.) This will enable us to determine the value of cr,kc_{r,k} and establish that Lemma 2.2 holds.

6 The case k=2k=2

We will first examine the case k=2k=2 in more detail. We must determine whether x=\nicefrac12x=\nicefrac{{1}}{{2}} is a local minimum or maximum of η\eta. If it is a local minimum, then it is the global minimum. Otherwise, there is a unique local minimum ξ∈(0,\nicefrac12)\xi\in(0,\nicefrac{{1}}{{2}}). To resolve this, we must examine η\eta in the neighbourhood of x=\nicefrac12x=\nicefrac{{1}}{{2}}. We show in Lemma 4.14 that \nicefrac12\nicefrac{{1}}{{2}} is a local minimum of η\eta for 2≤r≤42\leq r\leq 4, but is a local maximum if r≥5r\geq 5. Thus, for r=2, 3, 4r=2,\,3,\,4, the global minimum is ξ=\nicefrac12\xi=\nicefrac{{1}}{{2}}. (Note that we include the case r=k=2r=k=2 here, though ultimately it plays no part in our analysis.) Hence from (52) and (60) we have that for r=2, 3, 4r=2,\,3,\,4,

Now ur,1=0u_{r,1}=0 for all rr, and ur,2=2r−1ln⁡2u_{r,2}=2^{r-1}\ln 2, so

as required. (We cannot use this result in Theorem 1.1 when k=r=2k=r=2, since there is no sharp threshold in this case.)

In the cases k=2, r≥5k=2,\,r\geq 5, there is a local minimum ξ∈(0,\nicefrac12)\xi\in(0,\nicefrac{{1}}{{2}}), so the optimisation has similar characteristics to k≥3k\geq 3. We consider these cases in Section 3.8 below.

7 The case r=2r=2

We will consider the case r=2r=2 separately, since η\eta can be minimised exactly in this case. The results given in this section were obtained by Achlioptas and Naor in , by making a nonlinear substitution in η\eta. We can derive their results more simply, since we know that η\eta has a unique minimum. We have

Hence (63) implies that xx minimises η\eta if and only if

It is easily verified that x=1/kx=1/k satisfies this equation, and hence is the unique minimum of η\eta in [0,1−\nicefrac1k][0,1-\nicefrac{{1}}{{k}}].

We have dealt with the case k=2k=2 in the previous section, so we now assume that k≥3k\geq 3. Then

Now (k−1)3/k(k−2)=(k−1)(1+1/k(k−2))(k-1)^{3}/k(k-2)=(k-1)(1+1/k(k-2)) which lies strictly between k−1k-1 and kk, for k≥3k\geq 3. Thus

8 The general case

We now consider the remaining cases k≥3k\geq 3 or k=2, r≥5k=2,\,r\geq 5. We will do this by finding values w,y∈(0,1−\nicefrac1k)w,y\in(0,1-\nicefrac{{1}}{{k}}) such that η′(w)≤0\eta^{\prime}(w)\leq 0 and η′(y)>0\eta^{\prime}(y)>0. That is, ww satisfies (63), but yy does not. The uniqueness of ξ\xi then implies that w≤ξ<yw\leq\xi<y, and we will use this to place a lower bound on η(ξ)\eta(\xi). We will achieve this for all pairs r, kr,\,k except for a small number, and we will solve these few remaining cases numerically.

To simplify the analysis, we will exclude some cases initially. Thus we assume below that

First we set x=wx=w in (63), where w=(k−1)/krw=(k-1)/k^{r}. Note that w<1/r2w<1/r^{2}, from (72). Using Lemmas 4.5, 4.6 and 4.8, we have

and the right hand side is bounded below by 11 whenever

We may easily show that the left hand side of (74) is decreasing with rr for r≥3r\geq 3, and it is clearly decreasing with k≥2k\geq 2. The right hand side is independent of kk and increasing with rr. Now (74) holds by calculation when (k,r)∈{(2,5), (3,4), (4,3)}(k,r)\in\{(2,5),\,(3,4),\,(4,3)\}. Therefore (74) holds for all (k,r)(k,r) which satisfy (71), and combining this with (73) shows that ww satisfies (63), as desired.

We now set x=yx=y in (63), where y=(k+2)/kry=(k+2)/k^{r}. We have ry<1/rry<1/r from (72). Then, using Lemmas 4.5 and 4.6, we have

Now yr−2/(k−1)r−1<1y^{r-2}/(k-1)^{r-1}<1 for r,k≥2r,k\geq 2 and ry<1/r<\nicefrac12ry<1/r<\nicefrac{{1}}{{2}}, using Lemma 4.8. Therefore

So p′(y)>0p^{\prime}(y)>0 if yy does not satisfy (63); that is, if

Dividing by yy and rearranging gives the equivalent condition

From Lemma 4.15, we have r2y≤1r^{2}y\leq 1 and that y=(r2y)/r2y=(r^{2}y)/r^{2} is decreasing with both rr and kk. Since y<1y<1, it follows easily that (1+y)yr−2/(k−1)r−1(1+y)y^{r-2}/(k-1)^{r-1} is decreasing with rr and kk. We may now check numerically that (1+y)yr−2/(k−1)r−1≤\nicefrac150(1+y)y^{r-2}/(k-1)^{r-1}\leq\nicefrac{{1}}{{50}} for all k, rk,\,r satisfying (71). It follows that (75) is implied by the inequality

We show in Lemma 4.16 that, if (76) holds for some r≥3r\geq 3, k≥2k\geq 2, then it holds for any r′,k′r^{\prime},k^{\prime} such that r′≥rr^{\prime}\geq r, k′≥kk^{\prime}\geq k. We may verify numerically that (76) holds for the following pairs r,kr,k.

Thus it holds for all pairs r, kr,\,k such that

Let us call these the pairs (k,r)(k,r) regular, with the remaining nineteen pairs being irregular. We deal with the irregular pairs below by numerical methods.

First we continue our focus on regular pairs. For such pairs we have argued that (k−1)/kr≤ξ<(k+2)/kr(k-1)/k^{r}\leq\xi<(k+2)/k^{r} and hence, using Lemmas 4.7 and 4.9,

using Lemma 4.18. Now kr−1>(k−1)r−1+(r−1)(k−1)r−2≥(k−1)r−1+2k^{r-1}>(k-1)^{r-1}+(r-1)(k-1)^{r-2}\geq(k-1)^{r-1}+2 for r≥3, k≥2r\geq 3,\,k\geq 2, which shows that

as required, and this holds for all r, k≥2r,\,k\geq 2.

Next we consider irregular pairs and use (63) to bound ξ\xi numerically, by bisection. This is quite straightforward, since we know that ξ∈(0,1−\nicefrac1k)\xi\in(0,1-\nicefrac{{1}}{{k}}) is unique. The resulting values of cr,kc_{r,k} are shown below, along with the corresponding values of ur,k−1u_{r,k-1} and ur,ku_{r,k}.

By inspection, ur,k−1<cr,k<ur,ku_{r,k-1}<c_{r,k}<u_{r,k} for all irregular pairs.

We have already proved most of Lemma 2.2, and we complete the task below.

The above analysis shows the existence of constants cr,kc_{r,k} for all r,k≥2r,k\geq 2 such that FF has a unique maximum at J\boldsymbol{J} whenever c<cr,kc<c_{r,k}. Combining the numerical results for irregular pairs with (67), (69), (77) and (78) shows that cr,k∈(ur,k−1, ur,k)c_{r,k}\in(u_{r,k-1},\,u_{r,k}) for all r,k≥2r,k\geq 2.

It remains to prove that for all r,k≥2r,k\geq 2 we have

This follows from (64) if k=2k=2 and r=2,3,4r=2,3,4, or from (70) if r=2r=2 and k≥3k\geq 3. In all other cases we have cr,k<(kr−1−1)ln⁡kc_{r,k}<(k^{r-1}-1)\ln k, from (78). Furthermore, it follows from Lemma 4.20 that r(r−1)ln⁡k/(kr−1−1)<1r(r-1)\ln k/(k^{r-1}-1)<1 whenever k≥3k\geq 3, r≥2r\geq 2, or k=2k=2, r≥5r\geq 5. This completes the proof of Lemma 2.2. ∎

Combining this result with the conclusion of Section 2, we see that Theorem 1.1 is established.

9 Asymptotics

We have given precise bounds on cr,kc_{r,k}, but if we require only asymptotic estimates as r→∞r\to\infty and/or k→∞k\to\infty then the following simplified analysis suffices.

When we write “r→∞r\to\infty and/or k→∞k\to\infty”, this is not to be interpreted as “r(n)→∞r(n)\to\infty and/or k(n)→∞k(n)\to\infty”, but merely as “rr and/or kk are arbitrarily large constants”. Otherwise, we cannot use Theorem 1.3 to establish the existence of a sharp threshold between cr,kc_{r,k} and ur,ku_{r,k}. This is the approach to asymptotic estimates taken, for example, in .

We will use (48) to improve the estimate of cr,kc_{r,k} asymptotically, as discussed in Remarks 3.4 and 3.5. First let us consider the maximum possible improvement that we might be able to achieve.

From Remark 3.5, we know that the maximum value of zz in (19) cannot be smaller than that given by taking t=kt=k in (48). Thus we may bound the possible increase in cr,kc_{r,k} as follows. Since g(β)≤1−1/kr−1g(\beta)\leq 1-1/k^{r-1} and t≤kt\leq k, it follows from (49), using Lemma 4.5, that ϑ≤1/(kr−1−1)\vartheta\leq 1/(k^{r-1}-1) in (50). Thus ∑i=0∞ϑi/(i+1)!≤1+ϑ/2+O(ϑ2)\sum_{i=0}^{\infty}\vartheta^{i}/(i+1)!\leq 1+\vartheta/2+O(\vartheta^{2}) in (50). Therefore we can increase cr,kc_{r,k} asymptotically by a factor at most 1+1/(2kr−1)+O(1/k2r−2)1+1/(2k^{r-1})+O(1/k^{2r-2}). Since cr,k<ur,k=kr−1ln⁡kc_{r,k}<u_{r,k}=k^{r-1}\ln k, the additive improvement to cr,kc_{r,k} from fully optimising (19) is at most \nicefrac12ln⁡k+O(ln⁡k/kr−1)\nicefrac{{1}}{{2}}\ln k+O(\ln k/k^{r-1}). Hence we cannot improve cr,kc_{r,k} asymptotically by more than an additive term \nicefrac12ln⁡k\nicefrac{{1}}{{2}}\ln k.

Therefore, let φ\varphi be the function defined by

Thus φ(x)\varphi(x) is minimised at ξ^=(k−1)/kr∈R\hat{\xi}=(k-1)/k^{r}\in\mathcal{R}, as expected. We can write

Since κ=4ξ^\kappa=4\hat{\xi}, using (79) we have,

Therefore, since η\eta has a unique minimum in [0,1−\nicefrac1k][0,1-\nicefrac{{1}}{{k}}], we have

We have g(x)=1−O(r/kr−1)g(x)=1-O(r/k^{r-1}) when x≤κx\leq\kappa, and hence ϑ=2/kr−O(r/k2r−1)\vartheta=2/k^{r}-O(r/k^{2r-1}), taking t=2t=2 in (49). Thus the factor (eϑ−1)/ϑ(e^{\vartheta}-1)/\vartheta in (50) is 1+1/kr−O(r/k2r−1)1+1/k^{r}-O(r/k^{2r-1}). This is effectively the maximum value of (eϑ−1)/ϑ(e^{\vartheta}-1)/\vartheta for x∈[0,1−\nicefrac1k]x\in[0,1-\nicefrac{{1}}{{k}}] and (eϑ−1)/ϑ(e^{\vartheta}-1)/\vartheta is effectively constant for x≤κx\leq\kappa. Thus

for any k, r≥2k,\,r\geq 2, provided krk^{r} is large enough. Thus, after multiplying the right side of (81) by (kr−1−1)2/kr−1(k^{r-1}-1)^{2}/k^{r-1}, the additive improvement in cr,kc_{r,k} is ln⁡k/k−O(r2ln⁡k/kr−1)\ln k/k-O(r^{2}\ln k/k^{r-1}). Applying this to (80), we have

the result obtained by Achlioptas and Moore for 22-colouring rr-uniform hypergraphs. The case r=2r=2 (colouring random graphs), studied by Achlioptas and Naor , is discussed further below.

The best lower bound on ur,ku_{r,k} is u~r,k=ur,k−\nicefrac12ln⁡k\widetilde{u}_{r,k}=u_{r,k}-\nicefrac{{1}}{{2}}\ln k from Remark 2.1, so there is a gap

Asymptotically, this gap is always nonzero, though extremely small compared to cr,kc_{r,k} or ur,ku_{r,k}. It is independent of rr (up to the error term), and grows slowly with kk. It is minimised when k=2k=2 and r→∞r\to\infty. The existence of this gap merely indicates that the second moment method is not powerful enough to pinpoint the sharp threshold. We know from Theorem 1.3 that the threshold lies in [cr,k,u~r,k][c_{r,k},\widetilde{u}_{r,k}], although it is possible that it does not converge to a constant as n→∞n\to\infty. Note that if we could obtain the maximum possible correction \nicefrac12ln⁡k\nicefrac{{1}}{{2}}\ln k, as discussed above, then the gap would be approximately (k−1)/k(k-1)/k, and hence uniformly bounded for all k, r≥2k,\,r\geq 2 except k=r=2k=r=2.

Observe that the asymptotic estimate of cr,kc_{r,k} given in (82) is not sharp in one case, namely when r=2r=2 and k→∞k\to\infty. Here the error in (82) is O(ln⁡k/k)O(\ln k/k), so we have not improved (80). Since this is the important case of colouring random graphs, we will examine it separately.

From (68), we know that the bound on c2,kc_{2,k} from minimising η\eta is precisely

The right side of (82) is φ(ξ^)+O(ln⁡k/k)\varphi(\hat{\xi})+O(\ln k/k), so (81) still implies that, when kk is large enough, we need only consider ϑ(x)\vartheta(x) for x∈Rx\in\mathcal{R}. It follows, as above, that the factor (eϑ−1)/ϑ=1+1/k2−O(1/k3)(e^{\vartheta}-1)/\vartheta=1+1/k^{2}-O(1/k^{3}). Thus the additive improvement in c2,kc_{2,k} is ln⁡k/k−O(ln⁡k/k2)\ln k/k-O(\ln k/k^{2}). Adding this to (83), we have

which marginally improves (69) asymptotically. Note that, taken together, (82) and (84) exhaust the possibilities for the manner in which rr and/or kk can grow large.

References

Appendix: Technical lemmas

G∈G∗(n,r,cn)G\in\mathcal{G}^{*}(n,r,cn) has at least (k−1)(k-1) isolated vertices a.a.s..

Define m=⌊cn⌋m=\lfloor cn\rfloor and let Y(v)Y(\boldsymbol{v}) be the number of isolated vertices in GG, determined by v\boldsymbol{v}. The mrmr entries of v\boldsymbol{v} are uniform on [n][n], from which it follows that E[Y]=n(1−1/n)mr∼ne−cr\mathbf{E}[Y]=n(1-1/n)^{mr}\sim ne^{-cr}. Also, the entries of v\boldsymbol{v} are independent, and arbitrarily changing any single entry can only change Y(v)Y(\boldsymbol{v}) by ±1\pm 1. Thus we may apply a standard martingale inequality [13, Corollary 2.27] to give

for large nn. Thus GG has Ω(n)\Omega(n) isolated vertices a.a.s., from which the result follows easily. ∎

Suppose that MM is a p×pp\times p matrix of q×qq\times q blocks, such that

We have, by adding and subtracting rows and columns of MM,

We can use the same transformations to compute det⁡B\det B, replacing BB by the 1×11\times 1 unit matrix in the argument. We obtain det⁡B=(q+1) 1q−1=q+1\det B=(q+1)\,1^{q-1}=q+1. Hence det⁡M=(p+1)q(q+1)p\det M=(p+1)^{q}(q+1)^{p}. ∎

(We are grateful to Brendan McKay for pointing out that (p+1)q(q+1)p(p+1)^{q}(q+1)^{p} is the number of spanning trees in the complete bipartite graph Kp+1,q+1K_{p+1,q+1}. This suggests that an alternative proof of the above lemma may be possible using Kirchhoff’s Matrix Tree Theorem, but we do not explore this here.)

If ρ∈[1,kr−1]\rho\in[1,k^{r-1}] then the system defined by (19b)–(19e) is feasible.

Firstly, note that the system (19c)–(19e) defines a convex set. The k×kk\times k matrix J\boldsymbol{J} with all entries equal to 1/k1/k is feasible when ρ=1\rho=1, while any k×kk\times k permutation matrix Δ\boldsymbol{\Delta} is feasible when ρ=kr−1\rho=k^{r-1}. Now define the k×kk\times k matrices A(ϵ)=(1−ϵ)J0+ϵΔ\boldsymbol{A}(\epsilon)=(1-\epsilon)\boldsymbol{J}_{0}+\epsilon\boldsymbol{\Delta} for all ϵ∈\epsilon\in. Then A(ϵ)\boldsymbol{A}(\epsilon) satisfies (19c)–(19e) by convexity, while (19b) becomes

Now Ψ(ϵ)\Psi(\epsilon) is a polynomial function of ϵ\epsilon, and hence continuous. Also Ψ(0)=1\Psi(0)=1 and Ψ(1)=kr−1\Psi(1)=k^{r-1}. Therefore, by the Intermediate Value Theorem, for any ρ∈[1,kr−1]\rho\in[1,k^{r-1}] there is some ϵ∗∈\epsilon^{*}\in such that Ψ(ϵ∗)=ρ\Psi(\epsilon^{*})=\rho, and hence A(ϵ∗)\boldsymbol{A}(\epsilon^{*}) is a feasible solution. ∎

At the point b\boldsymbol{b}, note that ∂z/∂aij\partial z/\partial a_{ij} is finite for all bij>0b_{ij}>0 and +∞+\infty for all bij=0b_{ij}=0. Thus, for all small enough δ>0\delta>0, there is a ball BB with centre b\boldsymbol{b} and radius δ\delta, such that z(a)>z(b)z(\boldsymbol{a})>z(\boldsymbol{b}) for every point a∈B′\boldsymbol{a}\in B^{\prime}, where B′=B∩SoB^{\prime}=B\cap\mathcal{S}^{o}. Note that B′B^{\prime} is a convex set. So, we need only show that there is a point in S′∩B′S^{\prime}\cap B^{\prime}, since this will contradict the assumption that b\boldsymbol{b} is a local maximum of zz.

Let us write the points in S\mathcal{S} as u=(a11,a12,a1k)\boldsymbol{u}=(a_{11},a_{12},a_{1k}) if t=1t=1, or as u=(a11,a12,a21,a2k)\boldsymbol{u}=(a_{11},a_{12},a_{21},a_{2k}) if t=2t=2. Let

Then ui∈So\boldsymbol{u}_{i}\in\mathcal{S}^{o} and ∥ui−b∥≤3θ\|\boldsymbol{u}_{i}-\boldsymbol{b}\|\leq 3\theta for θ∈(0,1)\theta\in(0,1) and i=1,2i=1,2. Thus, for small enough θ\theta, ui∈B′\boldsymbol{u}_{i}\in B^{\prime} (i=1,2i=1,2). Also

for small enough (positive) θ\theta, since b11≥b12>0b_{11}\geq b_{12}>0 and r≥2r\geq 2. Thus Φ(u0)>Φ(b)=ρ^\Phi(\boldsymbol{u}_{0})>\Phi(\boldsymbol{b})=\widehat{\rho}. Similarly

for small enough θ\theta, since bt1>0b_{t1}>0 (t=1,2)(t=1,2) and r≥2r\geq 2. Thus Φ(u1)<Φ(b)=ρ^\Phi(\boldsymbol{u}_{1})<\Phi(\boldsymbol{b})=\widehat{\rho}.

Now consider the points uϵ=(1−ϵ)u0+ϵu1\boldsymbol{u}_{\epsilon}=(1-\epsilon)\boldsymbol{u}_{0}+\epsilon\boldsymbol{u}_{1}, for ϵ∈\epsilon\in. By convexity, uϵ∈B′\boldsymbol{u}_{\epsilon}\in B^{\prime} for all ϵ∈\epsilon\in. Also Φ(uϵ)\Phi(\boldsymbol{u}_{\epsilon}) is a polynomial function of ϵ\epsilon with Φ(u0)>ρ^\Phi(\boldsymbol{u}_{0})>\widehat{\rho} and Φ(u1)<ρ^\Phi(\boldsymbol{u}_{1})<\widehat{\rho}. Hence, by the Intermediate Value Theorem, there exists ϵ∗∈\epsilon^{*}\in such that Φ(uϵ∗)=ρ^\Phi(\boldsymbol{u}_{\epsilon^{*}})=\widehat{\rho}. Then uϵ∗\boldsymbol{u}_{\epsilon^{*}} is the required point in S′∩B′\mathcal{S}^{\prime}\cap B^{\prime}. ∎

Let ϕ(z)=z−ln⁡(1+z)\phi(z)=z-\ln(1+z), which is strictly convex on z>−1z>-1, since ln⁡(1+z)\ln(1+z) is strictly concave. Also ϕ′(z)=1−1/(1+z)\phi^{\prime}(z)=1-1/(1+z), so ϕ\phi is stationary at z=0z=0, and this must be its unique minimum. Since ϕ(0)=0\phi(0)=0, we have ϕ(z)≥0\phi(z)\geq 0 for all z>−1z>-1, and ϕ(z)>0\phi(z)>0 if z≠0z\neq 0. ∎

ln⁡(1−z)≥−3z/2\ln(1-z)\geq-3z/2 for all 0≤z≤\nicefrac120\leq z\leq\nicefrac{{1}}{{2}}.

Let ϕ(z)=ln⁡(1−z)+3z/2\phi(z)=\ln(1-z)+3z/2. Then ϕ\phi is strictly concave on [0,1)[0,1), since ln⁡(1−z)\ln(1-z) is strictly concave. Also ϕ′(z)=−1/(1−z)+\nicefrac32\phi^{\prime}(z)=-1/(1-z)+\nicefrac{{3}}{{2}}, so ϕ\phi is stationary at z=\nicefrac13z=\nicefrac{{1}}{{3}}, and this must be its unique maximum. Now ϕ(0)=0\phi(0)=0, and we may calculate ϕ(\nicefrac12)>0\phi(\nicefrac{{1}}{{2}})>0, so ϕ(z)>0\phi(z)>0 for 0<z≤\nicefrac120<z\leq\nicefrac{{1}}{{2}}. ∎

For all z∈(0,1)z\in(0,1), (1−z)ln⁡(1−z)>−z(1-z)\ln(1-z)>-z and (1−12z)ln⁡(1−z)<−z(1-\tfrac{1}{2}z)\ln(1-z)<-z.

1+z≤1/(1−z)≤1+z+2z2≤1+2z1+z\leq 1/(1-z)\leq 1+z+2z^{2}\leq 1+2z for all 0≤z≤\nicefrac120\leq z\leq\nicefrac{{1}}{{2}}.

The first inequality is equivalent to z2≥0z^{2}\geq 0 if z<1z<1. The second inequality is equivalent to z≤\nicefrac12z\leq\nicefrac{{1}}{{2}}. The third follows trivially from the second. ∎

Let ϕ1(z)=(1−z)p−1+pz\phi_{1}(z)=(1-z)^{p}-1+pz. Then ϕ1(0)=0\phi_{1}(0)=0 and ϕ1′(z)=p(1−(1−z)p−1)≥0\phi^{\prime}_{1}(z)=p(1-(1-z)^{p-1})\geq 0 if z∈z\in, giving the first inequality. Let ϕ2(z)=1−pz+12(pz)2−(1−z)p\phi_{2}(z)=1-pz+\tfrac{1}{2}(pz)^{2}-(1-z)^{p}. Then ϕ2(0)=0\phi_{2}(0)=0 and ϕ2′(z)=−p+p2z+p(1−z)p−1≥−p+p2z+p(1−(p−1)z)=pz≥0\phi^{\prime}_{2}(z)=-p+p^{2}z+p(1-z)^{p-1}\geq-p+p^{2}z+p(1-(p-1)z)=pz\geq 0, by the first inequality, giving the second. For the third inequality, using Lemma 4.5, we have (1−z)p≤e−pz=1/epz≤1/(1+pz)(1-z)^{p}\leq e^{-pz}=1/e^{pz}\leq 1/(1+pz). ∎

Let η(x)=f(x)/g(x)\eta(x)=f(x)/g(x) for x∈[0,1−1/k)x\in[0,1-1/k), where

Furthermore, if η′(x)=0\eta^{\prime}(x)=0 and g(x)≠0g(x)\neq 0 then ω(x)=η(x)\omega(x)=\eta(x).

The stated value of η(0)\eta(0) follows from the definition. Recall the calculations of Section 3.5. Using L’Hôpital’s rule ,

The same calculations prove that lim⁡x→1−1/kω(x)\lim_{x\to 1-1/k}\omega(x) also takes this value. Next,

As x→0x\to 0, all quantities in (87) are finite, except f′(x)→−∞f^{\prime}(x)\to-\infty. Since g(0)>0g(0)>0, we have η′(x)→−∞\eta^{\prime}(x)\to-\infty as x→0x\to 0. Note also that the last statement of the lemma follows from (87).

For the final calculation note that when x=1−\nicefrac1kx=1-\nicefrac{{1}}{{k}}, the numerator and denominator in the expression for η′(x)\eta^{\prime}(x) are both zero. Hence applying L’Hôpital’s rule again gives

using the values of f′′(1−\nicefrac1k)f^{\prime\prime}(1-\nicefrac{{1}}{{k}}), f′′′(1−\nicefrac1k)f^{\prime\prime\prime}(1-\nicefrac{{1}}{{k}}), g′′(1−\nicefrac1k)g^{\prime\prime}(1-\nicefrac{{1}}{{k}}) and g′′′(1−\nicefrac1k)g^{\prime\prime\prime}(1-\nicefrac{{1}}{{k}}) calculated in Section 3.5. ∎

Let k≥2k\geq 2 and r≥2r\geq 2 be integers, and let λ\lambda be a positive real number. If r≤2kr\leq 2k then the equation

has at most two solutions for xx in [0,1−\nicefrac1k][0,1-\nicefrac{{1}}{{k}}]. Otherwise r≥2k+1r\geq 2k+1 and the above equation has at most two solutions for xx in [0,1−\nicefrac1k][0,1-\nicefrac{{1}}{{k}}] whenever λ<λ0\lambda<\lambda_{0}, where

We summarise the behaviour of ϕ\phi in $inthefollowingtable.Herein the following table. Here\downarrowmeans“decreasing”,means “decreasing”,\uparrowmeans“increasing”.Thefinalcolumngivesthemaximumnumberofstationarypointsofmeans “increasing”. The final column gives the maximum number of stationary points of\phiinthecorrespondingsubintervalofin the corresponding subinterval of$.

We first consider small values of rr. When r=2r=2 the union of the first and last subinterval is ∖{\nicefrac12}\setminus\{\nicefrac{{1}}{{2}}\}, which contains no stationary point. Hence ϕ\phi has at most one stationary point in $(anditcanonlyoccurat(and it can only occur atx=\nicefrac{{1}}{{2}}).When). Whenr=3theunionofthefirst,secondandlastsubintervalequalsthe union of the first, second and last subinterval equals\setminus\{\nicefrac{{2}}{{3}}\}andcontainsatmostonestationarypoint.Henceand contains at most one stationary point. Hence\phihasatmosttwostationarypointsinhas at most two stationary points in$.

Next we assume that r≥5r\geq 5, which implies that all five subintervals are nonempty. Either ϕ\phi has one stationary point which is a local maximum, or it has three stationary points: a local maximum μ1\mu_{1}, a local minimum μ2\mu_{2}, and a local maximum μ3\mu_{3}, with μ1<μ2<μ3\mu_{1}<\mu_{2}<\mu_{3}.

(We take L2=−∞L_{2}=-\infty if there is only one stationary point.) First we show that

Next we calculate an upper bound on L2L_{2} by considering two cases. First, if 2≤r≤k2\leq r\leq k then

since θ(1−x)\theta(1-x) is maximised at x=1−1/rx=1-1/r in $.Next,if. Next, ifr>k\geq 2thenthen\theta(1-x)ismaximisedwhenis maximised whenx=1/kinin[0,1-\nicefrac{{1}}{{k}}]$. Therefore

Thus (88) holds if (r−2)2r−1+(r/2)r<(r−1)r−1(r-2)2^{r-1}+(r/2)^{r}<(r-1)^{r-1}. We show in Lemma 4.12 that this is true for all r≥5r\geq 5, so (88) holds for r≥5r\geq 5. Now ϕ\phi has at least one local maximum, so we have established that ϕ\phi has a local maximum μ1∈[1/r,2/r)\mu_{1}\in[1/r,2/r) whenever r≥5r\geq 5.

We now consider whether ϕ\phi has a local minimum μ2∈(2/r,1−2/r)\mu_{2}\in(2/r,1-2/r). Since there is a local maximum μ1∈[1/r,2/r)\mu_{1}\in[1/r,2/r) we know that ϕ′(2/r)<0\phi^{\prime}(2/r)<0. Thus ϕ\phi has a local minimum μ2∈(2/r,1−2/r]\mu_{2}\in(2/r,1-2/r] if and only if ϕ′(1−2/r)>0\phi^{\prime}(1-2/r)>0. Now

For all r≥5r\geq 5 the inequality (r−2)2r−1+(r/2)r<(r−1)r−1(r-2)2^{r-1}+(r/2)^{r}<(r-1)^{r-1} holds.

We will show (r−2)2r−1<(r−1)r−1/2(r-2)2^{r-1}<(r-1)^{r-1}/2 and (r/2)r<(r−1)r−1/2(r/2)^{r}<(r-1)^{r-1}/2.

To show (r−2)2r−1<(r−1)r−1/2(r-2)2^{r-1}<(r-1)^{r-1}/2, let γ1(r)=2(r−2)2r−1/(r−1)r−1\gamma_{1}(r)=2(r-2)2^{r-1}/(r-1)^{r-1}. Then

if r≥4r\geq 4. Thus γ1(r)\gamma_{1}(r) is decreasing for r≥4r\geq 4. Since γ1(5)=\nicefrac38<1\gamma_{1}(5)=\nicefrac{{3}}{{8}}<1, the inequality follows.

To show (r/2)r<(r−1)r−1/2(r/2)^{r}<(r-1)^{r-1}/2, let γ2(r)=rr/(2r−2)r−1\gamma_{2}(r)=r^{r}/(2r-2)^{r-1}. Then

if r≥4r\geq 4. Thus γ2(r)\gamma_{2}(r) is decreasing for r≥4r\geq 4. Since γ2(5)=55/212<1\gamma_{2}(5)=5^{5}/2^{12}<1, the inequality follows. ∎

(The values of η(0)\eta(0) and η(1−\nicefrac1k)\eta(1-\nicefrac{{1}}{{k}}) are stated in Lemma 4.10 while λ0\lambda_{0} is defined in Lemma 4.11.)

The left hand inequality reduces to r(r−1)ln⁡k/(kr−1−1)<1r(r-1)\ln k/(k^{r-1}-1)<1. In Lemma 4.20 we show that r(r−1)ln⁡k/(kr−1−1)<1r(r-1)\ln k/(k^{r-1}-1)<1 for all k≥3k\geq 3, r≥2r\geq 2, or k=2k=2, r≥5r\geq 5. Clearly this includes all r≥2k+1r\geq 2k+1, and so establishes the left hand inequality.

which is equivalent to γ(r,k)<1\gamma(r,k)<1, where

if r2(r−1)≤(r−2)(r+1)2r^{2}(r-1)\leq(r-2)(r+1)^{2}. This is equivalent to r2−3r−2≥0r^{2}-3r-2\geq 0, which is true for all r≥4r\geq 4. Thus γ(r,k)\gamma(r,k) is decreasing in rr, so we need only establish the critical case r=2k+1r=2k+1. We have

if (2k−2)(2k+1)2−(2k−1)(2k)2>0(2k-2)(2k+1)^{2}-(2k-1)(2k)^{2}>0, which is 2k2−3k−1>02k^{2}-3k-1>0. This holds for all k≥2k\geq 2. ∎

If k=2k=2 then x=\nicefrac12x=\nicefrac{{1}}{{2}} is a local minimum of η\eta for r=2, 3, 4r=2,\,3,\,4, and a local maximum if r≥5r\geq 5.

If r=2,3r=2,3 then the coefficient of z2z^{2} is positive, so z=0z=0 is a local minimum. If r≥5r\geq 5 then the coefficient of z2z^{2} is negative, so z=0z=0 is a local maximum. However, if r=4r=4, the coefficient of z2z^{2} is zero, so we need a higher order approximation. We compute

The coefficient of z4z^{4} is positive, and hence z=0z=0 is a local minimum. ∎

The function r2(k+2)/krr^{2}(k+2)/k^{r} is decreasing in both rr and kk for all r≥3, k≥2r\geq 3,\,k\geq 2. Hence r2(k+2)/kr<1r^{2}(k+2)/k^{r}<1 if

if k≥(1+1/r)2k\geq(1+1/r)^{2}. Since (1+1/r)2≤\nicefrac169(1+1/r)^{2}\leq\nicefrac{{16}}{{9}} for r≥3r\geq 3, this is satisfied for all k≥2k\geq 2. Also

We can now check numerically that r2(k+2)/kr<1r^{2}(k+2)/k^{r}<1 for (k,r)∈{(2,9), (3,4), (4,3)}(k,r)\in\{(2,9),\,(3,4),\,(4,3)\}. ∎

holds for some (r,k)(r,k) with r≥3r\geq 3, k≥2k\geq 2, then it holds for all (r′,k′)(r^{\prime},k^{\prime}) such that r′≥rr^{\prime}\geq r, k′≥kk^{\prime}\geq k.

The right side of this inequality is decreasing with rr and kk, so it suffices to show that the function ϕ(r,k)\phi(r,k) on the left side is increasing. This follows since, if k≥2, r≥3k\geq 2,\,r\geq 3,

since (k+1)(k+2)>k(k+3)(k+1)(k+2)>k(k+3) for all k≥0k\geq 0. Now we will have

if the function γ(x)=x/ln⁡x\gamma(x)=x/\ln x is increasing for x≥kx\geq k. Since γ′(x)=(ln⁡x−1)/(ln⁡x)2>0\gamma^{\prime}(x)=(\ln x-1)/(\ln x)^{2}>0 for x>ex>e, we have ϕ(r,k+1)/ϕ(r,k)>1\phi(r,k+1)/\phi(r,k)>1 for k≥3k\geq 3. For k=2k=2 and r≥3r\geq 3, we may verify that

Thus ϕ(r,k)\phi(r,k) is increasing in kk and rr for all k≥2, r≥3k\geq 2,\,r\geq 3, and the conclusion follows. ∎

For all regular pairs, r(rln⁡k+1)ξ≤1r(r\ln k+1)\xi\leq 1.

We have ξ≤(k+2)/kr\xi\leq(k+2)/k^{r} for all regular pairs. Thus the inequality is true if ϕ(r,k)≤1\phi(r,k)\leq 1, where ϕ(r,k)=r(rln⁡k+1)(k+2)/kr\phi(r,k)=r(r\ln k+1)(k+2)/k^{r}. Now

is decreasing with k≥2k\geq 2 for all r≥3r\geq 3, since ln⁡k/k2\ln k/k^{2} is decreasing for k≥2k\geq 2. Also

if r≥3, k≥2r\geq 3,\,k\geq 2. Thus ϕ(r,k)\phi(r,k) is decreasing with r≥3r\geq 3 for k≥2k\geq 2. Direct calculation shows that ϕ(9,2)\phi(9,2), ϕ(6,3)\phi(6,3), ϕ(5,4)\phi(5,4), ϕ(4,5)\phi(4,5), ϕ(3,15)\phi(3,15) are all less than 1. Thus r(rln⁡k+1)(k+2)/kr<1r(r\ln k+1)(k+2)/k^{r}<1 for all regular pairs. ∎

For all regular pairs, ln⁡k−2(k+2)/kr>ln⁡(k−1)\ln k-2(k+2)/k^{r}>\ln(k-1).

Using Lemma 4.5, ln⁡k−ln⁡(k−1)=−ln⁡(1−1/k)>1/k>2(k+2)/kr\ln k-\ln(k-1)=-\ln(1-1/k)>1/k>2(k+2)/k^{r}, provided 2+4/k<kr−22+4/k<k^{r-2}. The left hand side of 2+4/k<kr−22+4/k<k^{r-2} is decreasing, and the right hand side increasing, for all r, kr,\,k. Thus we need only determine the smallest pairs r≥3, k≥2r\geq 3,\,k\geq 2 which satisfy it. These are k=2, r=5k=2,\,r=5, k=3, r=4k=3,\,r=4 and k=4, r=3k=4,\,r=3, which are not regular pairs. ∎

For all k≥1k\geq 1, 4(k−1)≥kln⁡k4(k-1)\geq\sqrt{k}\ln k.

Using Lemma 4.5, ln⁡k=2ln⁡k≤2(k−1)\ln k=2\ln\sqrt{k}\leq 2(\sqrt{k}-1). So the conclusion is implied by 2(k−1)≥k−k2(k-1)\geq k-\sqrt{k}, which follows from 2(k−1)≥(k−1)2(k-1)\geq(k-1) for all k≥1k\geq 1. ∎

r(r−1)ln⁡k/(kr−1−1)<1r(r-1)\ln k/(k^{r-1}-1)<1 for all k≥3k\geq 3, r≥2r\geq 2, or k=2k=2, r≥5r\geq 5.

Let ϕ(r,k)=r(r−1)ln⁡k/(kr−1−1)\phi(r,k)=r(r-1)\ln k/(k^{r-1}-1). Then, for k≥3k\geq 3, r≥2r\geq 2, or k=2k=2, r≥5r\geq 5 we have

for k≥3k\geq 3, from the proof of Lemma 4.16. If k=2k=2, r≥5r\geq 5 then

So ϕ\phi is decreasing with both rr and kk. Now we may calculate that